WBCHSE Class 12 Maths Solutions For The Plane

WBCHSE Class 12 Maths Solutions For The Plane

Plane A plane is a surface such that a line segment joining any two points on it lies wholly on it.

Normal To A Plane A straight line which is perpendicular to every line lying on a plane is called a normal to the plane.

All the normals to a plane are parallel to each other.

General Equation of a Plane in the Cartesian Form

Theorem 1 Every equation ax + by + cz + d = 0 of the first degree in x, y, z always represents a plane. Also, a, b, c are the direction ratios of the normal to this plane.

Proof

Let us consider a surface represented by the equation

ax + by + cz + d = 0 …(1)

Let A(x₁, y₁, z₁) and B(x₂, y₂, z₂) be any two points on the surface represented by (1). Then,

ax₁ + by₁ + cz₁ + d = 0 …(2)

and, ax₂ + by₂ + cz₂ + d = 0 …(3)

Multiplying (3) by λ and adding to (2), we get

a(λx₂ + x₁) + b(λy₂ + y₁) + c(λz₂ + z₁) + d(λ+1) = 0

⇒ \(a\left(\frac{\lambda x_2+x_1}{\lambda+1}\right)+b\left(\frac{\lambda y_2+y_1}{\lambda+1}\right)+c\left(\frac{\lambda z_2+z_1}{\lambda+1}\right)+d=0\)

⇒ \(\left(\frac{\lambda x_2+x_1}{\lambda+1}, \frac{\lambda y_2+y_1}{\lambda+1}, \frac{\lambda z_2+z_1}{\lambda+1}\right)\) lies on surface (1), when λ ≠ -1.

But, these are the general coordinates of a point which divides AB in the ratio λ : 1.

Since λ may take any real value other than -1, it follows that every point of AB lies on (1).

Hence, ax + by + cz + d = 0 represents a plane.

Read and Learn More  Class 12 Math Solutions

To Show that a, b, c are the Direction Ratios of the Normal to a plane

Subtracting (2) from (3), we get

a(x₂ – x₁) + b(y₂ – y₁) + c(z₂ – z₁) = 0

⇒ a line with direction ratios a, b, c is perpendicular to an arbitrary line AB taken on plane (1) [∵ (x₂-x₁), (y₂-y₁), (z₂-z₁) are d.r.’s of AB]

⇒ a line with d.r.’s a, b, c is perpendicular to the plane (1)

⇒ a, b, c are the direction ratios of the normal to the plane (1).

Hence, ax + by + cz + d = 0 represents a plane, and a, b, c are the direction ratios of the normal to this plane.

Equation of a Plane Passing through a Given Point

Theorem 2 The equation of a plane passing through a point P(x₁,y₁,z₁) is a(x-x₁) + b(y-y₁) + c(z-z₁) = 0, where a, b, c are constants.

Proof

The general equation of a plane is

ax + by + cz + d = 0 …(1)

If this plane passes through the point P(x₁,y₁,z₁) then

ax₁ + by₁ + cz₁ + d = 0 …(2)

Subtracting (2) from (1), we get

a(x-x₁) + b(y-y₁) + c(z-z₁) = 0.

This is the general equation of a plane passing through the point P(x₁,y₁,z₁).

WBCHSE Class 12 Maths Solutions For The Plane Solved Examples

Example 1 Find the equation of the plane passing through the points A(2,3,4), B(-3,5,1) and C(4,-1,2).

Solution

Given 

A(2,3,4), B(-3,5,1) and C(4,-1,2)

The general equation of a plane passing through the point A(2,3,4) is given by

a(x-2) + b(y-3) + c(z-4) = 0 …(1)

Since it passes through the points B(-3,5,1) and C(4,-1,2), we have

a(-3-2) + b(5-3) + c(1-4) = 0 ⇒ -5a + 2b – 3c = 0 …(2)

a(4-2) + b(-1-3) + c(2-4) = 0 ⇒ 2a – 4b – 2c = 0 ⇒ a – 2b – c = 0 …(3)

Cross multiplying (2) and (3), we have

\(\frac{a}{(-2-6)}=\frac{b}{(-3-5)}=\frac{c}{(10-2)}\)

⇒ \(\frac{a}{-8}=\frac{b}{-8}=\frac{c}{8}\)

⇒ \(\frac{a}{1}=\frac{b}{1}=\frac{c}{-1}=k \text { (say) }\)

⇒ a = k, b = k and c = -k.

Substituting a = k, b = k and c = -k in (1), we get

k(x-2) + k(y-3) – k(z-4) = 0

⇒ (x-2) + (y-3) – (z-4) = 0

⇒ x + y – z – 1 = 0.

Hence, x + y – z = 1 is the required equation of the plane.

To Show that Four Given Points are Coplanar

Suppose, we have to show that four points A, B, C, D are coplanar. We proceed in the following steps:

1. Find the equation of the plane passing through any three (say A, B, C) of the given four points.

2. Show that the fourth point D satisfies the equation obtained in step 1.

Example 2 Show that the four points A(3,2,-5), B(-1,4,-3), C(-3,8,5) and D(-3,2,1) are coplanar. Find the equation of the plane containing them.

Solution

Given

A(3,2,-5), B(-1,4,-3), C(-3,8,5) and D(-3,2,1)

The equation of the plane passing through the point A(3,2,-5) is

a(x-3) + b(y-2) + c(z+5) = 0 …(1)

If it passes through B(-1,4,-3) and C(-3,8,-5), we have

a(-1-3) + b(4-2) + c(-3+5) = 0 ⇒ – 4a + 2b + 2c = 0 ⇒ 2a – b – c = 0 …(2)

a(-3-3) + b(8-2) + c(-5+5) = 0 ⇒ -6a + 6b + 0c = 0 ⇒ a – b – 0c = 0 …(3)

Cross multiplying (2) and (3), we get

\(\frac{a}{(0-1)}=\frac{b}{(-1-0)}=\frac{c}{(-2+1)}\)

⇒ \(\frac{a}{-1}=\frac{b}{-1}=\frac{c}{-1}\)

⇒ \(\frac{a}{1}=\frac{b}{1}=\frac{c}{1}=k \text { (say) }\)

⇒ a = k, b = k and c = k in (1), we get

k(x-3) + k(y-2) + k(z+5) = 0

⇒ (x-3) + (y-2) + (z+5) = 0 ⇒ x + y + z = 0.

Thus, the equation of the plane passing through the points A(3,2,-5), B(-1,4,-3) and C(-3,8,-5) is x + y + z = 0.

Clearly, the fourth point D(-3,2,1) also satisfies x + y + z = 0.

Hence, the given four points are coplanar, and the equation of the plane containing them is x + y + z = 0.

Equation of a Plane in the Intercept Form

Theorem 3 If a plane makes intercepts of lengths a, b, c with the x-axis, y-axis and z-axis respectively, the equation of the plane is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.\)

Proof

Let O be the origin, and let the plane meet the coordinate axes at A, B, C respectively such that OA = a, OB = b and OC = c.

So, the coordinates of these points are A(a,0,0), B(0,b,0) and C(0,0,c).

Let the equation of the given plane be

Ax + By + CZ + D = 0 …(1)

Since the given plane does not pass through O(0,0,0), D ≠ 0.

Also, since (1) passes through A(a,0,0), B(0,b,0) and C(0,0,c), we have

Aa + D = 0 ⇒ A = \(-\frac{D}{a}\),

Bb + D = 0 ⇒ B = \(-\frac{D}{b}\),

Cc + D = 0 ⇒ C = \(-\frac{D}{c}\).

Putting these values in (1), we get

\(\frac{-D x}{a}-\frac{D y}{b}-\frac{D z}{c}+D=0\)

⇒ \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) [on dividing throughout by -D].

This is the required equation of the plane in the intercept form.

Example 3 Find the equation of the plane which cuts off intercepts 6,3,-4 from the axes of coordinates.

Solution

We know that the equation of a plane which cuts off intercepts a, b, c from the x-axis, y-axis and z-axis respectively, is

\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.\)

Here, a = 6, b = 3 and c = -4.

Hence, the required equation of the plane is

\(\frac{x}{6}+\frac{y}{3}+\frac{z}{-4}=1 \Rightarrow 2 x+4 y-3 z=12.\)

Example 4 A variable plane moves in such a way that the sum of the reciprocals of its intercepts on the coordinate axes is constant. Show that the plane passes through a fixed point.

Solution

Let the equation of the variable plane be

\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) …(1)

Then, it makes intercepts a, b, c with the coordinate axes.

∴ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=k\), where k is a constant (given)

⇒ \(\frac{1}{k a}+\frac{1}{k b}+\frac{1}{k c}=1\)

⇒ \(\frac{1}{a}\left(\frac{1}{k}\right)+\frac{1}{b}\left(\frac{1}{k}\right)+\frac{1}{c}\left(\frac{1}{k}\right)=1\)

⇒ \(\left(\frac{1}{k}, \frac{1}{k}, \frac{1}{k}\right)\) satisifes (1).

Hence, the given plane passes through a fixed point \(\left(\frac{1}{k}, \frac{1}{k}, \frac{1}{k}\right)\).

Equation of a Plane in the Normal Form

Vector Form

Theorem 1 If \(\hat{n}\) is a unit vector normal to a given plane, directed from the origin to the plane, and p is the length of the perpendicular drawn from the origin to the plane then the vector equation of the plane is \(\vec{r} \cdot \hat{n}=p .\)

Proof

Let O be the origin, and let ON be the perpendicular drawn from O to the given plane. Let ON = p.

Let \(\hat{n}\) be a unit vector along \(\overrightarrow{O N}\).

Then, \(\overrightarrow{O N}=p \hat{n}\).

Let P be an arbitrary point on the plane, and let the position vector of P be \(\vec{r}\).

Then, \(\overrightarrow{O P}=\vec{r} \text {. }\)

Since \(\overrightarrow{N P}\) lies on the plane, \(\overrightarrow{N P}\) is perpendicular to \(\hat{n}\).

∴ \(\overrightarrow{N P} \cdot \hat{n}=0\)

⇒ \((\overrightarrow{O P}-\overrightarrow{O N}) \cdot \hat{n}=0\)

⇒ \((\vec{r}-p \hat{n}) \cdot \hat{n}=0\)

⇒ \(\vec{r} \cdot \hat{n}-p \hat{n} \cdot \hat{n}=0\)

⇒ \(\vec{r} \cdot \hat{n}=p .\)

Hence, the required equation of the plane is \(\vec{r} \cdot \hat{n}=p .\)

Remark The equation of a plane which is at a distance p from the origin and which is perpendicular to \(\hat{n}\) is \(\vec{r} \cdot \hat{n}=p .\)

Corollary If \(\vec{n}\) is a vector normal to a given plane then \(\vec{r} \cdot \vec{n}=q\) represents a plane.

Proof \(\vec{r} \cdot \vec{n}=q ⇒ \vec{r} \cdot \frac{\vec{n}}{|\vec{n}|}=\frac{q}{|\vec{n}|}\)

⇒ \(\vec{r} \cdot \hat{n}=p\), where \(\frac{q}{|\vec{n}|}=p .\)

But, \(\vec{r} \cdot \hat{n}=p\) represents a plane.

∴ \(\vec{r} \cdot \vec{n}=q\) also represents a plane.

Cartesian Form

Theorem 2 If a, b, c be the direction ratios of the normal to a given plane, and p be the length of the perpendicular drawn from the origin to the given plane then the equation of the plane is ax + by + cz = p.

Proof

We know that the vector equation of a plane in the normal form is

\(\vec{r} \cdot \hat{n}=p\) …(1)

Taking \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k} \text { and } \hat{n}=a \hat{i}+b \hat{j}+c \hat{k}\), equation (1) becomes

\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(a \hat{i}+b \hat{j}+c \hat{k})=p\)

⇒ ax + by + cz = p, which is the required equation of the plane.

Reduction of the General Form of the Equation of a Plane to the Normal Form

Let the general equation of the plane be ax + y + cz + d = 0. Then,

ax + by + cz + d = 0

⇒ -ax – by – cz = d …(1)

⇒ \(\frac{-a x}{\sqrt{a^2+b^2+c^2}}+\frac{-b y}{\sqrt{a^2+b^2+c^2}}+\frac{-c z}{\sqrt{a^2+b^2+c^2}}=\frac{d}{\sqrt{a^2+b^2+c^2}}\)

[on dividing (1) through out by \(\sqrt{(-a)^2+(-b)^2+(-c)^2}\)]

⇒ lx + my + nz = p, where

l = \(\frac{-a}{\sqrt{a^2+b^2+c^2}}, m=\frac{-b}{\sqrt{a^2+b^2+c^2}}\)

n = \(\frac{-c}{\sqrt{a^2+b^2+c^2}} \text { and } p=\frac{d}{\sqrt{a^2+b^2+c^2}}\).

Here l, m, n are the d.c.’s of the normal to the plane and p is the length of the perpendicular from the origin to the plane.

Solved Examples

Example 1 Find the vector equation of a plane which is at a distance of 6 units from the origin and which has \(\hat{j}\) as the unit vector normal to it.

Solution

Clearly, the required equation of the plane is \(\vec{r} \cdot \hat{j}=6 .\)

Example 2 Find the vector equation of a plane which is at a distance of 7 units from the origin and which is normal to the vector \((\hat{i}+2 \hat{j}-2 \hat{k})\)

Solution

Here, \(\vec{n}=(\hat{i}+2 \hat{j}-2 \hat{k})\) and p = 7.

∴ \(\hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{(\hat{i}+2 \hat{j}-2 \hat{k})}{\sqrt{1^2+2^2+(-2)^2}}=\left(\frac{1}{3} \hat{i}+\frac{2}{3} \hat{j}-\frac{2}{3} \hat{k}\right)\)

Hence, the required equation of the plane is

\(\vec{r} \cdot\left(\frac{1}{3} \hat{i}+\frac{2}{3} \hat{j}-\frac{2}{3} \hat{k}\right)=7 \text {, i.e., } \vec{r} \cdot(\hat{i}+2 \hat{j}-2 \hat{k})=21 \text {. }\)

Example 3 The vector equation of a plane is \(\vec{r} \cdot(2 \hat{i}-\hat{j}+2 \hat{k})=9\). Reduce it to the normal form, and hence find the length of the perpendicular from the origin to the plane.

Solution

The equation of the given plane is

\(\vec{r} \cdot \vec{n}=9 \text {, where } \vec{n}=(2 \hat{i}-\hat{j}+2 \hat{k}) \text {. }\)

∴ \(|\vec{n}|=\sqrt{2^2+(-1)^2+2^2}=3 .\)

Now, \(\vec{r} \cdot \vec{n}=9 \Rightarrow \vec{r} \cdot \frac{\vec{n}}{|\vec{n}|}=\frac{9}{|\vec{n}|}\)

⇒ \(\vec{r} \cdot \frac{(2 \hat{i}-\hat{j}+2 \hat{k})}{3}=3\)

⇒ \(\vec{r} \cdot\left(\frac{2}{3} \hat{i}-\frac{1}{3} \hat{j}+\frac{2}{3} \hat{k}\right)=3 .\)

Thus, \(\vec{r} \cdot\left(\frac{2}{3} \hat{i}-\frac{1}{3} \hat{j}+\frac{2}{3} \hat{k}\right)=3\) is the required equation of the plane in the normal form.

The length of the perpendicular from the origin to the plane is 3 units.

Example 4 Find a unit vector normal to the plane \(\vec{r} \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})+14=0 .\)

Solution

The equation of the given plane is

\(\vec{r} \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})+14=0\)

⇔ \(\vec{r} \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})=-14\)

⇒ \(\vec{r} \cdot(-2 \hat{i}+3 \hat{j}-6 \hat{k})=14\)

⇒ \(\vec{r} \cdot \vec{n}=14 \text {, where } \vec{n}=(-2 \hat{i}+3 \hat{j}-6 \hat{k})\)

⇔ \(\vec{r} \cdot \frac{\vec{n}}{|\vec{n}|}=\frac{14}{|\vec{n}|} \text {; where }|\vec{n}|=\sqrt{(-2)^2+3^2+(-6)^2}=7\)

⇒ \(\vec{r} \cdot \frac{(-2 \hat{i}+3 \hat{j}-6 \hat{k})}{7}=\frac{14}{7}\)

⇔ \(\vec{r} \cdot\left(-\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}-\frac{6}{7} \hat{k}\right)=2 \text {. }\)

Hence, the unit vector normal to the given plane is \(\hat{n}=\left(-\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}-\frac{6}{7} \hat{k}\right) .\)

Example 5 Find the direction cosines of the perpendicular from the origin to the plane \(\vec{r} \cdot(6 \hat{i}-3 \hat{j}-2 \hat{k})+3=0\).

Solution

The given equation may be written as \(\vec{r} \cdot(6 \hat{i}-3 \hat{j}-2 \hat{k})=-3\)

⇒ \(\vec{r} \cdot(-6 \hat{i}+3 \hat{j}+2 \hat{k})=3\)

⇒ \(\vec{r} \cdot \vec{n}=3 \text {, where } \vec{n}=(-6 \hat{i}+3 \hat{j}+2 \hat{k})\)

⇒ \(\vec{r} \cdot \frac{\vec{n}}{|\vec{n}|}=\frac{3}{|\vec{n}|} \text {, where }|\vec{n}|=\sqrt{(-6)^2+3^2+2^2}=7\)

⇒ \(\vec{r} \cdot \frac{(-6 \hat{i}+3 \hat{j}+2 \hat{k})}{7}=\frac{3}{7}\)

⇒ \(\vec{r} \cdot\left(-\frac{6}{7} \hat{i}+\frac{3}{7} \hat{j}+\frac{2}{7} \hat{k}\right)=\frac{3}{7} .\)

Hence, the direction cosines of the normal to the plane are \(\left(-\frac{6}{7}, \frac{3}{7}, \frac{2}{7}\right)\).

Example 6 Find the vector equation of a plane which is at a distance of 5 units from the origin and which has -1, 2, 2 as the direction ratios of a normal to it.

Solution

The direction ratios of the normal to the plane are -1, 2, 2. So, the normal vector is given by \(\hat{n}=-\hat{i}+2 \hat{j}+2 \hat{k}\).

∴ \(|\vec{n}|=\sqrt{(-1)^2+2^2+2^2}=3\)

⇒ \(\hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{(-\hat{i}+2 \hat{j}+2 \hat{k})}{3}=\left(-\frac{1}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\right) \text {. }\)

Hence, the required equation of the plane is

\(\vec{r} \cdot \hat{n}=5 \text {, i.e., } \vec{r} \cdot\left(-\frac{1}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\right)=5 \text {. }\)

Example 7 Find the Cartesian equation of a plane whose vector equation is \(\vec{r} \cdot(2 \hat{i}+5 \hat{j}-4 \hat{k})=3\).

Solution

We have

\(\vec{r} \cdot(2 \hat{i}+5 \hat{j}-4 \hat{k})=3\)

⇔ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}+5 \hat{j}-4 \hat{k})=3\)

⇔ 2x + 5y – 4z = 3.

Hence, the required equation is 2x + 5y – 4z = 3.

Example 8 Find the vector equation of a plane whose Cartesian equation is 5x – 7y + 2z = 4.

Solution

We have

5x – 7y + 2z = 4.

⇔ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(5 \hat{i}-7 \hat{j}+2 \hat{k})=4\)

⇔ \(\vec{r} \cdot(5 \hat{i}-7 \hat{j}+2 \hat{k})=4 .\)

Hence, the required equation is \(\vec{r} \cdot(5 \hat{i}-7 \hat{j}+2 \hat{k})=4 .\)

Example 9 Find a vector normal to the plane x + 2y + 3z – 6 = 0.

Solution

We have

x + 2y + 3z = 6

⇔ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+2 \hat{j}+3 \hat{k})=6\)

⇔ \(\hat{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})=6 .\)

Hence, the vector normal to the given plane is \((\hat{i}+2 \hat{j}+3 \hat{k})\).

Example 10 Reduce the equation of the plane x – 2y + 2z – 9 = 0 to the normal form, and hence find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane.

Solution

The equation of the given plane is x – 2y + 2z = 9 …(1)

Dividing throughout by \(\sqrt{1^2+(-2)^2+2^2}\), i.e., by 3, we get

\(\frac{1}{3} x-\frac{2}{3} y+\frac{2}{3} z=3\), which is the required normal form.

The length of the perpendicular from the origin to the plane = 3 units.

The direction cosines of the normal to the plane are \(\frac{1}{3}\), –\(\frac{2}{3}\), \(\frac{2}{3}\).

Example 11 Reduce the equation of the plane 12x – 3y + 4z + 5 = 0 to the normal form, and hence find the length of the perpendicular from the origin to the plane. Write down the direction cosines of the normal to the plane.

Solution

The given equation is

12x – 3y + 4z + 52 = 0

⇒ 12x – 3y + 4z = -52

⇒ -12x + 3y – 4z = 52 …(1)

Dividing (1) throughout by \(\sqrt{(-12)^2+3^2+(-4)^2}\), i.e., by 13, we get \(\left(\frac{-12}{13} x+\frac{3}{13} y-\frac{4}{13} z\right)=4\)

This is in the normal form, lx + my + nz = p.

The length of the perpendicular from the origin to this plane is 4 units.

The direction cosines of the normal to the plane are \(\left(\frac{-12}{13}, \frac{3}{13},-\frac{4}{13}\right)\) .

Example 12 The coordinates of the foot of the perpendicular drawn from the origin to a plane are (4, -2, -5). Find the equation of the plane.

Solution

Let O be the origin, and let N(4,-2,-5) be the foot of the perpendicular drawn from O to the given plane.

Let P(x,y,z) be an arbitrary point on the plane. Then, the d.r.’s of \(\overrightarrow{N P}\) are (x-4), (y+2), (z+5).

The d.r.’s of \(\overrightarrow{O N}\) are 4, -, -5.

Also, \(\overrightarrow{O N} \perp \overrightarrow{N P}\)

⇒ \(\overrightarrow{O N} \cdot \overrightarrow{N P}=0\)

⇒ 4(x-4) – 2(y+2) – 5(z+5) = 0

⇒ 4x – 2y – 5z – 45 = 0.

Hence, 4x – 2y – 5z = 45 is the required equation of the plane.

Example 13 Find the Cartesian form of the equation of the plane \(\vec{r}=(s-2 t) \hat{i}+(3-t) \hat{j}+(2 s+t) \hat{k} .\)

Solution

We have

\(\vec{r}=(s-2 t) \hat{i}+(3-t) \hat{j}+(2 s+t) \hat{k}\)

⇔ \((x \hat{i}+y \hat{j}+z \hat{k})=(s-2 t) \hat{i}+(3-t) \hat{j}+(2 s+t) \hat{k}\)

⇔ x = s – 2t, y = 3 – t and z = 2s + t

⇔ x – 2y = (s-6) and y + z = (3+2s) [eliminating t]

⇔ x – 2y + 6 = \(\frac{1}{2}\) (y + z – 3) [equating the values of s]

⇔ 2x – 4y + 12 = y + z – 3

⇔ 2x – 5y – z + 15 = 0.

This is the required Cartesian form of the equation of the given plane.

Equation of a Plane Parallel to a Given Plane

Vector Form

Any plane parallel to the plane \(\vec{r} \cdot \vec{n}=d_1\) is given by \(\vec{r} \cdot \vec{n}=d_2\), where the constant d2 is determined by a given condition.

Example Find the vector equation of the plane through (3,4,-1), which is parallel to the plane \(\vec{r} \cdot(2 \hat{i}-3 \hat{j}+5 \hat{k})+2=0 .\)

Solution

Let the required equation of the plane be

\(\vec{r} \cdot(2 \hat{i}-3 \hat{j}+5 \hat{k})=d\) …(1)

Since it passes through the point A(3,4,-1), which has the position vector \((3 \hat{i}+4 \hat{j}-\hat{k})\), we have

\((3 \hat{i}+4 \hat{j}-\hat{k}) \cdot(2 \hat{i}-3 \hat{j}+5 \hat{k})=d\)

⇔ d = [3 x 2 + 4 x (-3) + (-1) x 5] = -11.

Hence, the required equation of the plane is

\(\vec{r} \cdot(2 \hat{i}-3 \hat{j}+5 \hat{k})+11=0 .\)

Cartesian Form

Let ax + by + cz + d = 0 be a given plane.

Then, any plane parallel to this plane is of the form ax + by + cz + k = 0, where k is determined by a given condition.

Example Find the equation of the plane which is parallel to the plane 2x – 3y + z + 8 = 0 and which passes through the point (-1,1,2).

Solution

Any plane parallel to the given plane is 2x – 3y + z + k = 0 …(1)

If it passes through (-1,1,2) then

2 x (-1) – 3 x 1 + 2 + k = 0 ⇔ k = 3.

Hence, the required equation of the plane is 2x – 3y + z + 3 = 0.

Equation of a Plane through the Intersection of Two Planes

Vector Form

Theorem 1 The equation of a plane through the intersection of two planes \(\vec{r} \cdot \overrightarrow{n_1}=q_1 and \vec{r} \cdot \overrightarrow{n_2}=q_2\) is given by \(\vec{r} \cdot\left(\overrightarrow{n_1}+\lambda \overrightarrow{n_2}\right)=q_1+\lambda q_2.\)

Proof Let π₁ and π₂ be two intersecting planes, and let their equations be \(\vec{r} \cdot \overrightarrow{n_1}=q_1 and \vec{r} \cdot \overrightarrow{n_2}=q_2.\)

Let \(\overrightarrow{r_1}\) be the position vector of a point common to them.

Then, \(\vec{r} \cdot \overrightarrow{n_1}=q_1 and \vec{r} \cdot \overrightarrow{n_2}=q_2\)

⇒ \(\overrightarrow{r_1} \cdot \overrightarrow{n_1}+\lambda\left(\overrightarrow{r_1} \cdot \overrightarrow{n_2}\right)=q_1+\lambda q_2\)

⇒ \(\overrightarrow{r_1} \cdot\left(\overrightarrow{n_1}+\lambda \overrightarrow{n_2}\right)=q_1+\lambda q_2 .\)

Thus, \(\vec{r}=\overrightarrow{r_1}\) satisfies the equation \(\vec{r} \cdot\left(\overrightarrow{n_1}+\lambda \overrightarrow{n_2}\right)=q_1+\lambda q_2\).

Hence, the equation of any plane through the intersection of two planes \(\vec{r} \cdot \overrightarrow{n_1}=q_1 and \vec{r} \cdot \overrightarrow{n_2}=q_2\) is given by

\(\vec{r} \cdot\left(\overrightarrow{n_1}+\lambda \overrightarrow{n_2}\right)=q_1+\lambda q_2\).

Theorem 2 Two Planes always intersect in a line

Proof

Let π₁ and π₂ be two intersecting planes, and let \(\vec{a}\) be the position vector of a point common to them.

Then, their equations are of the form

\((\vec{r}-\vec{a}) \cdot \overrightarrow{n_1}=0 \text { and }(\vec{r}-\vec{a}) \cdot \overrightarrow{n_2}=0\)

⇒ \((\vec{r}-\vec{a})\) is perpendicular to each one of \(\overrightarrow{n_1} \text { and } \overrightarrow{n_2}\)

⇒ \((\vec{r}-\vec{a}) is parallel to \left(\overrightarrow{n_1} \times \overrightarrow{n_2}\right)\)

⇒ \((\vec{r}-\vec{a})=\lambda\left(\vec{n}_1 \times \vec{n}_2\right)\) for some scalar λ

⇒ \(\vec{r}=\vec{a}+\lambda \vec{b}, \text { where } \vec{b}=\left(\overrightarrow{n_1} \times \overrightarrow{n_2}\right) \text {. }\)

But, we know that \(\vec{r}=\vec{a}+\lambda \vec{b}\) is the vector equation of a line.

Hence, two planes alwalys intersect in a line.

Cartesian Form

Theorem 3 The equation of a plane through the intersection of two planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given by (a₁x + b₁y + c₁z + d₁) + λ(a₂x + b₂y + c₂z + d₂) = 0.

Proof

The given equations are

a₁x + b₁y + c₁z + d₁ = 0 …(1)

a₂x + b₂y + c₂z + d₂ = 0 …(2)

Consider the equation

(a₁x + b₁y + c₁z + d₁) + λ(a₂x + b₂y + c₂z + d₂) = 0

⇒ (a₁ + λa₂)x + (b₁ + λb₂)y + (c₁ + λc₂)z + (d₁ + λd₂) = 0 …(3)

This being a first – degree equation in x,y,z, represents a plane. In order to prove that (3) represents a plane passing through the intersection of the planes (1) and (2), it is sufficient to show that every point on the line of intersection of (1) and (2) is point on (3).

Let (α, β, γ) be a point common to (1) and (2).

Then, a₁α + b₁β + c₁γ + d₁ = 0 and a₂α + b₂β + c₂γ + d₂ = 0

⇒ (a₁α + b₁β + c₁γ + d₁) + λ(a₂α + b₂β + c₂γ + d₂) = 0

⇒ (α, β, γ)

⇒ (a₁α + b₁β + c₁γ + d₁) + λ(a₂α + b₂β + c₂γ + d₂) = 0 represents a plane through the intersection of the planes

a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0

Solved Examples

Example 1 Find the equation of the plane passing through the line of intersection of the planes 2x – 7y + 4z = 3 and 3x – 5y + 4z + 11 = 0 and the point (-2,1,3).

Solution

Any plane through the intersection of the two given plane is

(2x – 7y + 4z – 3) + λ(3x – 5y + 4z + 11) = 0 …(1)

If it passes through the point (-2,1,3), we have

[2 x (-2) – 7 x 1 + 4 x 3 – 3] + λ[3 x (-2) – 5 x 1 + 4 x 3 + 11] = 0 …(1)

⇔ -2 + 12λ = 0 ⇔ 12λ = 2 ⇔ λ = \(\frac{2}{12}\) = \(\frac{1}{6}\).

Putting λ = \(\frac{1}{6}\) in (1), we get the required equation of the plane as

(2x – 7y + 4z – 3) + \(\frac{1}{6}\) (3x – 5y + 4z + 11) = 0

⇔ (12x – 42y + 24z – 18) + (3x – 5y + 4z + 11) = 0

⇔ 15x – 47y + 28z – 7 = 0.

The equation of the plane passing through the line of intersection of the planes 15x – 47y + 28z – 7 = 0.

Example 2 Find the equation of the plane which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 and which is perpendicular to the plane 5x + 3y – 6z + 8 = 0.

Solution

Any plane through the intersection of the two given planes is given by

(x + 2y + 3z – 4) + λ(2x + y – z + 5) = 0

⇔ (1+2λ)x + (2+λ)y + (3-λ)z + (5λ-4) = 0 …(1)

If the plane (1) is perpendicular to the plane 5x + 3y – 6z + 8 = 0, we have

5(1+2λ) + 3(2+λ) – 6(3-λ) = 0 ⇔ λ = \(\frac{7}{19}\).

Hence, the required equation of the plane is

\(\left(1+\frac{14}{19}\right) x+\left(2+\frac{7}{19}\right) y+\left(3-\frac{7}{19}\right) z+\left(\frac{35}{19}-4\right)=0\)

⇔ \(\frac{33 x}{19}+\frac{45 y}{19}+\frac{50 z}{19}-\frac{41}{19}=0\) ⇔ 33x + 45y + 50z – 41 = 0.

The equation of the plane which contains the line of intersection of the planes 33x + 45y + 50z – 41 = 0.

Example 3 Find the distance between the parallel planes 2x – y + 3z + 4 = 0 and 6x – 3y + 9z – 3 = 0.

Solution

Let P(x₁,y₁,z₁) be any point on the plane 2x – y + 3z + 4 = 0.

Then, 2x₁ – y₁ + 3z₁ + 4 = 0 ⇒ 2x₁ – y₁ + 3z₁ = -4 …(1)

∴ the distance p of the point P(x₁,y₁,z₁) from the plane 6x – 3y + 9z = 0 is given by

P = \(\frac{\left|6 x_1-3 y_1+9 z_1-3\right|}{\sqrt{6^2+(-3)^2+9^2}}=\frac{\left|3\left(2 x_1-y_1+3 z_1\right)-3\right|}{\sqrt{126}}\)

= \(\frac{|3 \times(-4)-3|}{\sqrt{126}}=\frac{15}{3 \sqrt{14}}=\frac{5}{\sqrt{14}} \text { units. }\)

The distance between the parallel planes = \(\frac{|3 \times(-4)-3|}{\sqrt{126}}=\frac{15}{3 \sqrt{14}}=\frac{5}{\sqrt{14}} \text { units. }\)

Example 4 Find the equation of a plane through the intersection of the planes \(\vec{r} \cdot(\hat{i}+3 \hat{j}-\hat{k})=5 \text { and } \vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=3\), and passing through the point (2,1,-2).

Solution

Any plane through the intersection of the two given planes is

\([\vec{r} \cdot(\hat{i}+3 \hat{j}-\hat{k})-5]+\lambda[\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})-3]=0\)

⇒ \(\hat{r} \cdot[(1+2 \lambda) \hat{i}+(3-\lambda) \hat{j}+(\lambda-1) \hat{k}]-(5+3 \lambda)=0\) …(1)

If (1) passes through (2,1,-2) then \(\hat{r}=(2 \hat{i}+\hat{j}-2 \hat{k})\) should satisfy it.

∴ \((2 \hat{i}+\hat{j}-2 \hat{k}) \cdot[(1+2 \lambda) \hat{i}+(3-\lambda) \hat{j}+(\lambda-1) \hat{k}]-(5+3 \lambda)=0\)

⇔ 2(1+2λ) + (3-λ) – 2(λ-1) – (5+3λ) = 0

⇔ 2λ = 2 ⇔ λ = 1.

Putting λ = 1 in (1), we get the required equation of the plane as \(\vec{r} \cdot(3 \hat{i}+2 \hat{j})=8 .\)

Example 5 Find the vector equation of the plane through the line of intersection of the planes \(\vec{r} \cdot(2 \hat{i}-3 \hat{j}+4 \hat{k})=1 \text { and } \vec{r} \cdot(\hat{i}-\hat{j})+4=0\), and perpendicular to the plane \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})+8=0 .\)

Solution

Any plane through the line of intersection of the two given plane is

\([\vec{r} \cdot(2 \hat{i}-3 \hat{j}+4 \hat{k})-1]+\lambda[\vec{r} \cdot(\hat{i}-\hat{j})+4]=0\)

⇒ \(\vec{r} \cdot[(2+\lambda) \hat{i}-(3+\lambda) \hat{j}+4 \hat{k}]=1-4 \lambda\) …(1)

If this plane is perpendicular to the plane \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})+8=0\), we have

2(2+λ) + (3+λ) + 4 = 0 ⇔ 3λ + 11 = 0 ⇔ λ = \(-\frac{11}{3}\).

Putting λ = \(-\frac{11}{3}\) in (1), we get the required equation of the plane as

\(\vec{r} \cdot(-5 \hat{i}+2 \hat{j}+12 \hat{k})=47.\)

Example 6 Find the Cartesian as well as the vector equation of the plane through the intersection of the planes \(\vec{r} \cdot(2 \hat{i}+6 \hat{j})+12=0 \text { and } \vec{r} \cdot(3 \hat{i}-\hat{j}+4 \hat{k})=0\), which is at a unit distance from the origin.

Solution

Any plane through the intersection of the two given planes is given by

\([\vec{r} \cdot(2 \hat{i}+6 \hat{j})+12]+\lambda[\vec{r} \cdot(3 \hat{i}-\hat{j}+4 \hat{k})]=0\)

⇔ \(\vec{r} \cdot[(2+3 \lambda) \hat{i}+(6-\lambda) \hat{j}+(4 \lambda) \hat{k}]+12=0\) …(1)

This plane is at a unit distance from the origin.

∴ the length of the perpendicular from the origin to (1) is 1.

∴ \(\frac{12}{\sqrt{(2+3 \lambda)^2+(6-\lambda)^2+(4 \lambda)^2}}=1\)

⇔ (2+3λ)² + (6-λ)² + (4λ)² = 144

⇔ 26λ² + 40 = 144 ⇔ 26λ² = 104

⇔ λ² = 4 ⇔ λ = ±2.

Putting λ = 2 and λ = -2 in (1), we get the required equations as

\(\vec{r} \cdot(8 \hat{i}+4 \hat{j}+8 \hat{k})+12=0 \text {, and } \vec{r} \cdot(-4 \hat{i}+8 \hat{j}-8 \hat{k})+12=0\)

⇔ \(\vec{r} \cdot(2 \hat{i}+\hat{j}+2 \hat{k})+3=0 \text {, and } \vec{r} \cdot(-\hat{i}+2 \hat{j}-2 \hat{k})+3=0 \text {. }\)

In the Cartesian form, these equations are

\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}+\hat{j}+2 \hat{k})+3=0\), and

\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(-\hat{i}+2 \hat{j}-2 \hat{k})+3=0\)

⇔ 2x + y + 2z + 3 = 0 and -x + 2y – 2z + 3 = 0

⇔ 2x + y + 2z + 3 = 0 and x – 2y + 2z – 3 = 0.

Equation of a Plane Passing through a Given Point and Perpendicular to a Given Vector

Vector Form

Theorem 1 The vector equation of a plane passing through a point A with position vector \(\vec{a}\) and perpendicular to a given vector \(\vec{n}\) is \((\vec{r}-\vec{a}) \cdot \vec{n}=0 .\)

Proof

Let O be the origin, and let π be the given plane.

Let A be a given point on the plane with position vector \(\vec{a}\).

Let P be an arbitrary point on the plane with position vector \(\vec{r}\).

Then, \(\overrightarrow{O A}=\vec{a} \text {, and } \overrightarrow{O P}=\vec{r} \text {. }\)

∴ \(\overrightarrow{A P}=(\overrightarrow{O P}-\overrightarrow{O A})=(\vec{r}-\vec{a})\)

Let \(\overrightarrow{O N}=\vec{n}\) be normal to the plane.

Now, \(\overrightarrow{A P}\) lies in the plane and \vec{n} is normal to the plane.

∴ \(\overrightarrow{A P} \perp \vec{n}\)

⇒ \(\overrightarrow{A P} \cdot \vec{n}=0\)

⇒ \((\vec{r}-\vec{a}) \cdot \vec{n}=0 .\)

Hence, the equation of the plane passing through the point with position vector \vec{a} and perpendicular to \(\vec{n}\) is \((\vec{r}-\vec{a}) \cdot \vec{n}=0 .\)

Corollary The vector equation of the plane passing through the origin and perpendicular to \(\vec{n}\) is \(\vec{r} \cdot \vec{n}=0\).

Cartesian Form

Theorem 2 The Cartesian equation of the plane passing through a point A(x₁,y₁,z₁) and perpendicular to a line having direction ratios n1, n2, n3 is (x-x₁)n₁ + (y-y₁)n₂ + (z-z₁)n₃ = 0.

Proof

Let \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}, \vec{a}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k} \text { and } \vec{n}=n_1 \hat{i}+n_2 \hat{j}+n_3 \hat{k}\)

Then, \((\vec{r}-\vec{a}) \cdot \vec{n}=0\)

⇒ (x-x₁)n₁ + (y-y₁)n₂ + (z-z₁)n₃ = 0.

Equation of a Plane Passing through a Given Point and Parallel to Two Given Lines

Vector Form

Theorem 1 The vector equation of a plane passing through a given point with position vector \(\vec{a}\) and parallel to two given vector \(\vec{b}\) and \(\vec{c}\) is \((\vec{r}-\vec{a}) \cdot(\vec{b} \times \vec{c})=0 .\)

Proof

The required plane is parallel to vectors \(\vec{b}\) and \(\vec{c}\).

So, the vector \(\vec{n}=(\vec{b} \times \vec{c})\) is perpendicular to this plane.

Thus, we have to find the equation of a plane passing through the point with position vector \(\vec{a}\) and perpendicular to the vector \(\vec{n}\).

So, its equation is

\((\vec{r}-\vec{a}) \cdot \vec{n}=0 \text {, i.e., }(\vec{r}-\vec{a}) \cdot(\vec{b} \times \vec{c})=0 \text {. }\)

Hence, the required equation of the plane is \((\vec{r}-\vec{a}) \cdot(\vec{b} \times \vec{c})=0 .\)

Parametric Form

Theorem 2 The vector equation of a plane passing through a given point with position vector \(\vec{a}\) and parallel to two given vectors \(\vec{b}\) and \(\vec{c}\) is \(\vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c}\), where λ and μ are scalars.

Proof

Let A be a given point with position vector \(\vec{a}\). Let \(\vec{r}\) be the position vector of a arbitrary point P on the plane passing through the point A, and parallel to \(\vec{b}\) and \(\vec{c}\).

∴ \(\overrightarrow{A P}=(\text { p.v. of } P)-(\text { p.v. of } A)=(\vec{r}-\vec{a})\)

Clearly, the vectors \((\vec{r}-\vec{a})\), \(\vec{b} \text { and } \vec{c}\) are coplanar.

∴ \((\vec{r}-\vec{a})\) lies in the plane of \(\vec{b}\) and \(\vec{c}\).

So, \((\vec{r}-\vec{a})\) can be expressed as a linear combination of \(\vec{b}\) and \(\vec{c}\).

∴ there exist scalars λ and μ such that

\((\vec{r}-\vec{a})=\lambda \vec{b}+\mu \vec{c} \Rightarrow \vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c}\),

which is the required equation of the plane.

Cartesian Form

Theorem 3 The equation of the plane passing through a given point A(x₁,y₁,z₁) and parallel to two given lines having direction ratios b₁, b₂, b₃ and c₁, c₂, c₃ is

\(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|=0 .\)

Proof

Let us consider a plane passing through a given point A(x₁,y₁,z₁) and parallel to two given lines having direction ratios b₁, b₂, b₃ and c₁, c₂, c₃.

Let P(x,y,z) be an arbitrary point on the plane. Then,

\(\overrightarrow{A P}\) = (p.v. of P) – (p.v. of A)

= \((x \hat{i}+y \hat{j}+z \hat{k})-\left(x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\right)\)

= \(\left(x-x_1\right) \hat{i}+\left(y-y_1\right) \hat{j}+\left(z-z_1\right) \hat{k} .\)

It is given that the plane is parallel to two lines having direction ratios b₁, b₂, b₃ and c₁, c₂, c₃.

So, the given plane is parallel to each of the vectors

\(\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k} \text { and } \vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}\)

∴ \(\overrightarrow{A P}\), \(\vec{b}\) and \(\vec{c}\) are coplanar, and therefore their scalar triple product must be zero.

\(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|=0 .\)

This is the required equation of the plane passing through the point (x₁,y₁,z₁) and parallel to two given lines having direction ratios b₁, b₂, b₃ and c₁, c₂, c₃.

Equation of a Plane Passing through Three Given Points

Vector Form

Theorem 1 The vector equation of a plane passing through three noncollinear points with position vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) is \((\vec{r}-\vec{a}) \cdot[(\vec{b}-\vec{a}) \times(\vec{c}-\vec{a})]=0.\)

Proof

Let A, B, C be three given noncollinear points having position vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) respectively.

Let P be an arbitrary point on the plane passing through the points

A, B, C and let \(\vec{r}\) be the position vector of P. Then,

\(\overrightarrow{A P}=\text { (p.v. of } P \text { ) – (p.v. of } A)=(\vec{r}-\vec{a})\),

\(\overrightarrow{A B}=\text { (p.v. of } B \text { ) – (p.v. of } A)=(\vec{b}-\vec{a})\),

\(\overrightarrow{A C}=\text { (p.v. of } C \text { ) – (p.v. of } A)=(\vec{c}-\vec{a})\).

Since the points A, B, C, P lie on the plane, the vectors \(\overrightarrow{A P}\), \(\overrightarrow{A B}\), \(\overrightarrow{A C}\) are coplanar.

So, the scalar triple product of these vectors is 0.

∴ \([\overrightarrow{A P} \overrightarrow{A B} \overrightarrow{A C}]=0\)

⇒ \(\overrightarrow{A P} \cdot(\overrightarrow{A B} \times \overrightarrow{A C})=0\)

⇒ \((\vec{r}-\vec{a}) \cdot[(\vec{b}-\vec{a}) \times(\vec{c}-\vec{a})]=0\), which is the required equation of the plane.

Cartesian Form

Theorem 2 The equation of a plane passing through three given noncollinear points A(x₁,y₁,z₁), B(x₂,y₂,z₂) and C(x₃,y₃,z₃) is given by

\(\left|\begin{array}{lll}
x-x_1 & y-y_1 & z-z_1 \\
x-x_2 & y-y_2 & z-z_2 \\
x-x_3 & y-y_3 & z-z_3
\end{array}\right|=0 .\)

Proof

Let P(x,y,z) be an arbitrary point on the plane passing through three given noncollinear points A(x₁,y₁,z₁), B(x₂,y₂,z₂) and C(x₃,y₃,z₃). Then,

\(\overrightarrow{A P}=\text { (p.v. of } P \text { ) – (p.v. of } A)\)

= \((x \hat{i}+y \hat{j}+z \hat{k})-\left(x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\right)\)

= \(\left(x-x_1\right) \hat{i}+\left(y-y_1\right) \hat{j}+\left(z-z_1\right) \hat{k}\);

\(\overrightarrow{B P}=\text { (p.v. of } P \text { ) – (p.v. of } B)\)

= \((x \hat{i}+y \hat{j}+z \hat{k})-\left(x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\right)\)

= \(\left(x-x_2\right) \hat{i}+\left(y-y_2\right) \hat{j}+\left(z-z_2\right) \hat{k}\);

and \(\overrightarrow{C P}=\text { (p.v. of } P \text { ) – (p.v. of } C)\)

= \((x \hat{i}+y \hat{j}+z \hat{k})-\left(x_3 \hat{i}+y_3 \hat{j}+z_3 \hat{k}\right)\)

= \(\left(x-x_3\right) \hat{i}+\left(y-y_3\right) \hat{j}+\left(z-z_3\right) \hat{k}\)

Now, the vectors \(\overrightarrow{A P}\), \(\overrightarrow{B P}\), \(\overrightarrow{C P}\) being coplanar, their scalar triple product must be zero.

∴ \([\overrightarrow{A P} \overrightarrow{B P} \overrightarrow{C P}]=0\)

⇒ \(\left|\begin{array}{lll}
x-x_1 & y-y_1 & z-z_1 \\
x-x_2 & y-y_2 & z-z_2 \\
x-x_3 & y-y_3 & z-z_3
\end{array}\right|=0 .\) which is the required equation of the plane.

Angle between Two Planes

The angle between two given planes is the angle between their normals.

Vector Form

Theorem 1 If θ be the angle between two planes \(\vec{r} \cdot \overrightarrow{n_1}=q_1 \text { and } \vec{r} \cdot \overrightarrow{n_2}=q_2\)

then cos θ = \(\frac{\overrightarrow{n_1} \cdot \overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|} .\)

Proof

Let θ be the angle between two planes whose vector equations are \(\vec{r} \cdot \overrightarrow{n_1}=q_1 \text { and } \vec{r} \cdot \overrightarrow{n_2}=q_2.\)

Then, θ is the angle between \(\overrightarrow{n_1} and \overrightarrow{n_2}\).

∴ cos θ = \(\frac{\overrightarrow{n_1} \cdot \overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|} .\)

Remark The acute angle θ between θ is given by

cos θ = \(\frac{\overrightarrow{n_1} \cdot \overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|} .\)

Two Important Results

1. Condition for two planes to be perpendicular to each other

Two planes \(\vec{r} \cdot \overrightarrow{n_1}=q_1 \text { and } \vec{r} \cdot \overrightarrow{n_2}=q_2\) are perpendicular to each other

⇔ \(\overrightarrow{n_1} \perp \overrightarrow{n_2}\)

⇔ \(\overrightarrow{n_1} \cdot \overrightarrow{n_2}=0 .\)

2. Condition for two planes to be parallel to each other

Two planes \(\vec{r} \cdot \overrightarrow{n_1}=q_1 \text { and } \vec{r} \cdot \overrightarrow{n_2}=q_2\) are parallel to each other

⇔ \(\overrightarrow{n_1} \| \overrightarrow{n_2}\)

⇔ \(\overrightarrow{n_1}=\lambda \overrightarrow{n_2}\) for some scalar λ.

Remark 1 Any plane parallel to \(\vec{r} \cdot \vec{n}=q \text { is } \vec{r} \cdot \vec{n}=q_1 \text {. }\)

Remark 2 The equation of the plane parallel to the plane \(\vec{r} \cdot \vec{n}=q\) and passing through the point with position vector \(\vec{a}\) is \((\vec{r}-\vec{a}) \cdot \vec{n}=0 .\)

Cartesian Form

Theorem 2 Let θ be the angle between two planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0.

Then, cos θ = \(\frac{\left(a_1 a_2+b_1 b_2+c_1 c_2\right)}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)} \sqrt{\left(a_2^2+b_2^2+c_2^2\right)}} .\)

Proof

Let θ be the angle between the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0

Then, θ is the angle between their normals.

But, the direction ratios of the normals to the given planes are a₁, b₁, c₁ and a₂, b₂, c₂.

∴ cos θ = \(\frac{\left(a_1 a_2+b_1 b_2+c_1 c_2\right)}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)} \sqrt{\left(a_2^2+b_2^2+c_2^2\right)}} .\)

Remark

The acute angle θ between the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0

is given by cos θ = \(\frac{\left(a_1 a_2+b_1 b_2+c_1 c_2\right)}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)} \sqrt{\left(a_2^2+b_2^2+c_2^2\right)}} .\)

Two Important Results

1. Condition for two planes to be perpendicular to each other

Two planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are perpendicular to each other

⇔ their normals are perpendicular to each other

⇔ lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular to each other

⇔ a₁a₂ + b₁b₂ + c₁c₂ = 0.

2. Condition for two planes to be parallel to each other

Two planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are parallel to each other

⇔ their normals are parallel to each other

⇔ lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are parallel to each other

⇔ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} .\)

Remark 1 The equation of any plane parallel to the plane ax + by + cz + d = 0 is ax + by +cz + λ = 0

Remark 2 The equation of a plane passing through (x₁,y₁,z₁) and parallel to the plane ax + by + cz + d = 0 given by a(x-x₁) + b(y-y₁) + c(z-z₁) = 0.

Remark 3 The equation of any plane parallel to the xy-plane is z = λ.

This plane is perpendicular to the z-axis

Similarly, the equation of any plane parallel to the zx-plane is y = λ.

And, the equation of any plane parallel to the yz-plane is x = λ.

Solved Examples

Example 1 Find the angle between the planes \(\vec{r} \cdot(\hat{i}+\hat{j}+2 \hat{k})=5 \text { and } \vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=8 \text {. }\)

Solution

We know that the angle between the planes \(\vec{r} \cdot \overrightarrow{n_1}=q_1 \text { and } \vec{r} \cdot \overrightarrow{n_2}=q_2\) is given by

\(\cos \theta=\frac{\overrightarrow{n_1} \cdot \overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|} .\)

Here, \(\overrightarrow{n_1}=(\hat{i}+\hat{j}+2 \hat{k}) \text { and } \overrightarrow{n_2}=(2 \hat{i}-\hat{j}+\hat{k}) \text {. }\)

∴ \(\cos \theta=\frac{(\hat{i}+\hat{j}+2 \hat{k}) \cdot(2 \hat{i}-\hat{j}+\hat{k})}{|\hat{i}+\hat{j}+2 \hat{k}||2 \hat{i}-\hat{j}+\hat{k}|}\)

= \(\frac{\{1 \times 2+1 \times(-1)+2 \times 1\}}{\left\{\sqrt{1^2+1^2+2^2}\right\}\left\{\sqrt{2^2+(-1)^2+1^2}\right\}}=\frac{3}{6}=\frac{1}{2}\)

⇒ \(\theta=\frac{\pi}{3} .\)

Hence, the angle between the given planes is \(\left(\frac{\pi}{3}\right)\).

Example 2 Find the angle between the planes \(\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=5 \text { and } \vec{r} \cdot(3 \hat{i}-3 \hat{j}+5 \hat{k})=6\)

Solution

We know that the angle between the planes \(\vec{r} \cdot \overrightarrow{n_1}=q_1 \text { and } \vec{r} \cdot \overrightarrow{n_2}=q_2\) is given by

\(\cos \theta=\frac{\overrightarrow{n_1} \cdot \overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|} .\)

Here, \(\overrightarrow{n_1}=(2 \hat{i}+2 \hat{j}-3 \hat{k}) \text { and } \overrightarrow{n_2}=(3 \hat{i}-3 \hat{j}+5 \hat{k}) \text {. }\)

∴ \(\cos \theta=\frac{(2 \hat{i}+2 \hat{j}-3 \hat{k}) \cdot(3 \hat{i}-3 \hat{j}+5 \hat{k})}{|2 \hat{i}+2 \hat{j}-3 \hat{k}||3 \hat{i}-3 \hat{j}+5 \hat{k}|}\)

= \(\frac{\{2 \times 3+2 \times(-3)+(-3) \times 5\}}{\left\{\sqrt{2^2+2^2+(-3)^2}\right\}\left\{\sqrt{3^2+(-3)^2+5^2}\right\}}\)

= \(\frac{-15}{(\sqrt{17})(\sqrt{43})}=\frac{-15}{\sqrt{731}}\)

⇒ \(\theta=\cos ^{-1}\left(\frac{-15}{\sqrt{731}}\right)\)

This is the obtuse angle between the given planes.

The acute angle between the given planes is \(\cos ^{-1}\left(\frac{-15}{\sqrt{731}}\right)\).

Example 3 Find the value of λ for which the planes \(\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})=13 \text { and } \vec{r} \cdot(\lambda \hat{i}+2 \hat{j}-7 \hat{k})=9\) are perpendicular to each other.

Solution

We know that the planes \(\vec{r} \cdot \overrightarrow{n_1}=q_1 \text { and } \vec{r} \cdot \overrightarrow{n_2}=q_2\) are perpendicular to each other only when \(\overrightarrow{n_1} \cdot \overrightarrow{n_2}=0 .\)

Here, \(\overrightarrow{n_1}=(\hat{i}+2 \hat{j}+3 \hat{k}) \text { and } \overrightarrow{n_2}=(\lambda \hat{i}+2 \hat{j}-7 \hat{k}) \text {. }\)

∴ the given planes are perpendicular to each other

⇔ \(\overrightarrow{n_1} \cdot \overrightarrow{n_2}=0 .\)

⇔ \((\hat{i}+2 \hat{j}+3 \hat{k}) \cdot(\lambda \hat{i}+2 \hat{j}-7 \hat{k})=0\)

⇔ 1 x λ + 2 x 2 + 3 x (-7) = 0 ⇔ λ = 17.

Hence, the required value of λ is 17.

Example 4 Find the angle between the planes x + y + 2z = 9 and 2x – y + z = 6.

Solution

We know that the angle between the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given by

\(\cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\left(\sqrt{a_1^2+b_1^2+c_1^2}\right)\left(\sqrt{a_2^2+b_2^2+c_2^2}\right)} .\)

Here, a₁ = 1, b₁ = 1, c₁ = 2, and a₂ = 2, b₂ = -1, c₂ = 1.

∴ \(\cos \theta=\frac{1 \times 2+1 \times(-1)+2 \times 1}{\left(\sqrt{1^2+1^2+2^2}\right)\left\{\sqrt{2^2+(-1)^2+1^2}\right\}}=\frac{3}{(\sqrt{6})(\sqrt{6})}=\frac{3}{6}=\frac{1}{2}\)

⇒ \(\theta=\frac{\pi}{3} .\)

Hence, the angel between the given planes is \(\left(\frac{\pi}{3}\right)\).

Example 5 Find the angle between the planes 2x – 3y + 4z = 1 and -x + y = 4.

Solution

We know that the angle between the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given by

\(\cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\left(\sqrt{a_1^2+b_1^2+c_1^2}\right)\left(\sqrt{a_2^2+b_2^2+c_2^2}\right)} .\)

Here, a₁ = 2, b₁ = -3, c₁ = 4, and a₂ = -1, b₂ = 1, c₂ = 0.

∴ \(\cos \theta=\frac{2 \times(-1)+(-3) \times 1+4 \times 0}{\left\{\sqrt{2^2+(-3)^2+4^2}\right\}\left\{\sqrt{(-1)^2+1^2+0^2}\right\}}\)

= \(\frac{-5}{(\sqrt{29})(\sqrt{2})}=\frac{-5}{\sqrt{58}}\)

⇒ \(\theta=\cos ^{-1}\left(\frac{-5}{\sqrt{58}}\right)\)

The acute angle between the given planes is \(\cos ^{-1}\left(\frac{-5}{\sqrt{58}}\right)\).

Example 6 Find the value of λ for which the planes 2x – 4y + 3z = 7 and x + 2y + λz = 18 are perpendicular to each other.

Solution

We know that the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are perpendicular to each other only when a₁a₂ + b₁b₂ + c₁c₂ = 0.

Here, a₁ = 2, b₁ = -4, c₁ = 3 and a₂ = 1, b₂ = 2, c₂ = λ.

∴ the given planes are perpendicular to each other

⇔ 2 x 1 + (-4) x 2 + 3 x λ = 0

⇔ 3λ = 6 ⇔ λ = 2.

Hence, λ = 2.

Example 7 Find the equation of the plane passing through the point (1,4,-2) and parallel to the plane -2x + y – 3z = 0.

Solution

The equation of a plane parallel to the given plane is of the form -2x + y – 3z = k for some scalar k.

Since it passes through (1,4,-2), we have

-2 x 1 + 4 – 3 x (-2) = k ⇔ k = 8.

Hence, the required equation of the plane is -2x + y – 3z = 8.

Example 8 Find the equation of the plane passing through the point (1,1,1) and perpendicular to each of the following planes x + 2y + 3z = 7 and 2x – 3y + 4z = 0.

Solution

Any plane through (1,1,1) is a(x1) + b(y-1) + c(z-1) = 0 …(1)

Now, (1) being perpendicular to each of the planes x + 2y + 3z = 7 and 2x – 3y + 4z = 0, we have

a x 1 + b x 2 + c x 3 = 0 ⇒ a + 2b + 3c = 0 …(2)

a x 2 + b x (-3) + c x 4 ⇒ 2a – 3b + 4c = 0 …(3)

Cross multiplying (2) and (3), we get

\(\frac{a}{(8+9)}=\frac{b}{(6-4)}=\frac{c}{(-3-4)}=k \text { (say) }\)

⇒ \(\frac{a}{17}=\frac{b}{2}=\frac{c}{-7}=k\)

⇒ a = 17k, b = 2k and c = -7k.

Putting these values in (1), we get

17k(x-1) + 2k(y-1) – 7k(z-1) = 0

⇔ 17(x-1) + 2(y-1) – 7(z-1) = 0

⇔ 17x + 2y – 7z – 12 = 0, which is the required equation of the plane.

Example 9 Find the equation of the plane passing through the points (2,2,1) and (9,3,6), and perpendicular to the plane 2x + 6y + 6z = 1.

Solution

Let the required plane passing through (2,2,1) be

a(x-) + b(y-2) + c(z-1) = 0 …(1)

Since it also passes through the point (9,3,6), we have

a(9-2) + b(3-2) + c(6-1) = 0 ⇒ 7a + b + 5c = 0 …(2)

Also, (1) being perpendicular to the plane 2x + 6y + 6z = 1, we have

2a + 6b + 6c = 0 …(3)

Cross multiplying (2) and (3), we get

\(\frac{a}{(6-30)}=\frac{b}{(10-42)}=\frac{c}{(42-2)}\)

⇒ \(\frac{a}{-24}=\frac{b}{-32}=\frac{c}{40}\)

⇒ \(\frac{a}{3}=\frac{b}{4}=\frac{c}{-5}=k \text { (say) }\)

⇒ a = 3k, b = 4k, and c = -5k

Putting these values of a, b, c in (1), we get the required equation of the plane as

3k(x-2) + 4k(y-2) – 5k(z-1) = 0

⇔ 3(x-2) + 4(y-2) – 5(z-1) = 0

⇒ 3x + 4y – 5z – 9 = 0.

Example 10 Find the equation of the plane passing through the points (2,3,4), (-3,5,1) and (4,-1,2).

Solution

The equation of the plane passing through the point (2,3,4) is a(x-2) + b(y-3) + c(z-4) = 0 …(1)

If this plane passes through the points (-3,5,1) and (4,-1,2) then

a(-3-2) + b(5-3) + c(1-4) = 0 ⇒ -5a + 2b – 3c = 0 …(2)

a(4-2) + b(-1-3) + c(2-4) = 0 ⇒ 2a – 4b – 2c = 0 …(3)

Cross multiplying (2) and (3), we get

\(\frac{a}{(-4-12)}=\frac{b}{(-6-10)}=\frac{c}{(20-4)} \Rightarrow \frac{a}{-16}=\frac{b}{-16}=\frac{c}{16}\)

⇒ \(\frac{a}{1}=\frac{b}{1}=\frac{c}{-1}=k \text { (say) }\)

⇒ a = k, b = k, c = -k.

Putting a = k, b = k, c = -k in (1), we get the required equation of the plane as

k(x-2) + k(y-3) – k(z-4) = 0

⇔ (x-2) + (y-3) – (z-4) = 0

⇔ x + y – z – 1 = 0.

Hence, the required equation of the plane is x + y – z – 1 = 0.

Angle between a Line and a Plane

The angle between a line and a plane is the complement of the angle between the line and the normal to the plane.

Vector Form

Theorem 1 If θ be the angle between the line \(\vec{r}=\vec{a}+\lambda \vec{b}\) and the plane \(vec{r} \cdot \vec{n}=q\) then \(\sin \theta=\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|} .\)

Proof

Clearly, the line \(\vec{r}=\vec{a}+\lambda \vec{b}\) is parallel to \(\vec{b}\), and the plane \(\vec{r} \cdot \vec{n}=q\) is normal to \(\vec{n}\).

Let θ be the angle between the given line and the given plane.

Then, the angle between \(\vec{b}\) and \(\vec{n}\) is \(\left(\frac{\pi}{2}-\theta\right)\).

∴ \(\cos \left(\frac{\pi}{2}-\theta\right)=\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|} \Rightarrow \sin \theta=\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)

Two Important Results

1. Condition for a given line to be perpendicular to a given plane

The line \(\vec{r}=\vec{a}+\lambda \vec{b}\) is perpendicular to the plane \(\vec{r} \cdot \vec{n}=q\)

⇔ \(\vec{b}\) is perpendicular to the plane \(\vec{r} \cdot \vec{n}=q\)

⇔ \(\vec{b}\) is parallel to the normal \(\vec{n}\) to the plane

⇔ \(\vec{b}=t \vec{n}\), for some scalar t.

2. Condition for a given line to be parallel to a given plane

The line \(\vec{r}=\vec{a}+\lambda \vec{b}\) is parallel to the plane \(\vec{r} \cdot \vec{n}=q\)

⇔ \(\vec{b}\) is parallel to the plane \(\vec{r} \cdot \vec{n}=\)

⇔ \(\vec{b}\) is perpendicular to the normal \(\vec{n}\) to the plane

⇔ \(\vec{b} \cdot \vec{n}=0 .\)

Cartesian Form

Theorem 2 If θ is the angle between the line \(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\) and the plane a₂x + b₂y + c₂z + d = 0 then

\(\sin \theta=\frac{\left(a_1 a_2+b_1 b_2+c_1 c_2\right)}{\left(\sqrt{a_1^2+b_1^2+c_1^2}\right)\left(\sqrt{a_2^2+b_2^2+c_2^2}\right)} .\)

Proof

The direction ratios of the given line are a₁, b₁, c₁.

So, the given line is parallel to \(\vec{b} = a_1 \hat{i}+b_1 \hat{j}+c_1 \hat{k}\)

The normal to the given plane is parallel to \(\vec{n} = a_2 \hat{i}+b_2 \hat{j}+c_2 \hat{k}\).

Let θ be the angle between the given line and the given plane. Then, \(\left(\frac{\pi}{2}-\theta\right)\) is the angle between the given line and the normal to the given plane.

∴ \(\left(\frac{\pi}{2}-\theta\right) is the angle between \vec{b} and \vec{n}\)

⇒ \(\cos \left(\frac{\pi}{2}-\theta\right)=\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)

⇒ \(\sin \theta=\frac{\left(a_1 \hat{i}+b_1 \hat{j}+c_1 \hat{k}\right) \cdot\left(a_2 \hat{i}+b_2 \hat{j}+c_2 \hat{k}\right)}{\left|a_1 \hat{i}+b_1 \hat{j}+c_1 \hat{k}\right| \cdot\left|a_2 \hat{i}+b_2 \hat{j}+c_2 \hat{k}\right|}\)

⇒ \(\sin \theta=\frac{\left(a_1 a_2+b_1 b_2+c_1 c_2\right)}{\left(\sqrt{a_1^2+b_1^2+c_1^2}\right)\left(\sqrt{a_2^2+b_2^2+c_2^2}\right)} .\)

Important Results

1. Condition for the given line to be perpendicular to the given plane

The line \(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\) is perpendicular to the plane a₂x + b₂y + c₂z + d = 0

⇔ the line \(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\) is parallel to the normal to thr plane a₂x + b₂y + c₂z + d = 0

⇔ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} .\)

2. Condition for the given line to be parallel to the given plane

The line \(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\) is parallel to the plane a₂x + b₂y + c₂z + d = 0

⇔ the line \(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\) is perpendicular to the normal to the plane a₂x + b₂y + c₂z + d = 0

⇔ a₁a₂ + b₁b₂ + c₁c₂ = 0.

3. Distance between a line and a plane, parallel to each other

If the line \(\vec{r}=\vec{a}+\lambda \vec{b}\) is parallel to the plane \(\vec{r} \cdot \vec{n}=q\) then the distance between them is \(\frac{|\vec{a} \cdot \vec{n}-q|}{|\vec{n}|}\), which is the same as the distance of a point from the plane.

Summary

1. If θ is the angle between the line \(\vec{r}=\vec{a}+\lambda \vec{b}\) and the plane \(\vec{r} \cdot \vec{n}=q\) then

\(\sin \theta=\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)

2. The line \(\vec{r}=\vec{a}+\lambda \vec{b}\) is perpendicular to the plane \(\vec{r} \cdot \vec{n}=q\) only when \(\vec{b}=t \vec{n}\) for some scalar t.

3. (1) The line \(\vec{r}=\vec{a}+\lambda \vec{b}\) is parallel to the plane \(\vec{r} \cdot \vec{n}=q\) only when \(\vec{b} \cdot \vec{n}=0 .\)

(2) If the line \(\vec{r}=\vec{a}+\lambda \vec{b}\) is parallel to the plane \(\vec{r} \cdot \vec{n}=q\) then the distance between them is

\(\frac{|\vec{a} \cdot \vec{n}-q|}{|\vec{n}|}\)

(4) If θ is the angle between the line \(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\) and the plane a₂x + b₂y + c₂z + d = 0 then

\(\sin \theta=\frac{\left(a_1 a_2+b_1 b_2+c_1 c_2\right)}{\left(\sqrt{a_1^2+b_1^2+c_1^2}\right)\left(\sqrt{a_2^2+b_2^2+c_2^2}\right)} .\)

5. The line \(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\) is perpendicular to the plane a₂x + b₂y + c₂z + d = 0 only if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

6. The line \(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\) is parallel to the plane a₂x + b₂y + c₂z + d = 0 only if a₁a₂ + b₁b₂ + c₁c₂ = 0.

Solved Examples

Example 1 Find the angle between the line \(\vec{r}=(\hat{i}+\hat{j}-3 \hat{k})+\lambda(2 \hat{i}+2 \hat{j}+\hat{k})\) and the plane \(\vec{r} \cdot(6 \hat{i}-3 \hat{j}+2 \hat{k})=5 \text {. }\)

Solution

We know that the angle θ between the line \(\vec{r}=\vec{a}+\lambda \vec{b}\) and the plane \(\vec{r} \cdot \vec{n}=q\) is given by

\(\sin \theta=\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\)

Here, \(\vec{b}=2 \hat{i}+2 \hat{j}+\hat{k} \text { and } \vec{n}=6 \hat{i}-3 \hat{j}+2 \hat{k} \text {. }\)

∴ \(\sin \theta=\frac{(2 \hat{i}+2 \hat{j}+\hat{k}) \cdot(6 \hat{i}-3 \hat{j}+2 \hat{k})}{|2 \hat{i}+2 \hat{j}+\hat{k}||6 \hat{i}-3 \hat{j}+2 \hat{k}|}\)

= \(\frac{\{2 \times 6+2 \times(-3)+1 \times 2\}}{\left(\sqrt{2^2+2^2+1^2}\right)\left\{\sqrt{\left.6^2+(-3)^2+2^2\right\}}\right.}=\left(\frac{8}{3 \times 7}\right)=\frac{8}{21}\)

⇒ \(\theta=\sin ^{-1}\left(\frac{8}{21}\right) .\)

Hence, the angle between the given line and the given plane is \(\sin ^{-1}\left(\frac{8}{21}\right) .\)

Example 2 Find the value of m for which the line \(\vec{r}=(\hat{i}+2 \hat{j}-\hat{k})+\lambda(2 \hat{i}+\hat{j}+2 \hat{k})\) is parallel to the plane \(\vec{r} \cdot(3 \hat{i}-2 \hat{j}+m \hat{k})=12\).

Solution

We know that the line \(\vec{r}=\vec{a}+\lambda \vec{b}\) is parallel to the plane

\(\vec{r} \cdot \vec{n}=q \Leftrightarrow \vec{b} \cdot \vec{n}=0 .\)

Here, \(\vec{b}=2 \hat{i}+\hat{j}+2 \hat{k} \text { and } \vec{n}=3 \hat{i}-2 \hat{j}+m \hat{k} \text {. }\)

∴ the given line is parallel to the given plane

⇔ \(\vec{b} \cdot \vec{n}=0\)

⇔ \((2 \hat{i}+\hat{j}+2 \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+m \hat{k})=0\)

⇔ 2 x 3 + 1 x (-2) + 2 x m = 0

⇔ 2m = -4 ⇔ m = -2.

Hence, m = -2.

Example 3 Show that the line \(\vec{r}=(2 \hat{i}-2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+4 \hat{k})\) is parallel to the plane \(\vec{r} \cdot(\hat{i}+5 \hat{j}+\hat{k})=5\) . Also, find the distance between them.

Solution

We know that the line \(\vec{r}=\vec{a}+\lambda \vec{b}\) is parallel to the plane \(\vec{r} \cdot \vec{n}=q\) only when \(\vec{b} \cdot \vec{n}=0.\)

And, the distance between the line and the plane is given by

\(\frac{|\vec{a} \cdot \vec{n}-q|}{|\vec{n}|}\)

Here, \(\vec{a}=2 \hat{i}-2 \hat{j}+3 \hat{k}, \vec{b}=\hat{i}-\hat{j}+4 \hat{k}, \vec{n}=\hat{i}+5 \hat{j}+\hat{k} \text { and } q=5 \text {. }\)

Now, \(\vec{b} \cdot \vec{n}=(\hat{i}-\hat{j}+4 \hat{k}) \cdot(\hat{i}+5 \hat{j}+\hat{k})=\{1 \times 1+(-1) \times 5+4 \times 1\}=0\)

Hence, the given line is parallel to the given plane.

Distance between the given line and the given plane

= \(\frac{|\vec{a} \cdot \vec{n}-q|}{|\vec{n}|}=\frac{|(2 \hat{i}-2 \hat{j}+3 \hat{k}) \cdot(\hat{i}+5 \hat{j}+\hat{k})-5|}{|\hat{i}+5 \hat{j}+\hat{k}|}\)

= \(\frac{|2 \times 1+(-2) \times 5+3 \times 1-5|}{\sqrt{1^2+5^2+1^2}}\)

= \(\frac{|2-10+3-5|}{\sqrt{27}}=\frac{10}{3 \sqrt{3}} \text { units. }\)

Example 4 Find the vector equation of a line passing through the point with position vector \((2 \hat{i}-3 \hat{j}-5 \hat{k})\) and perpendicular to the plane \(\vec{r} \cdot(6 \hat{i}-3 \hat{j}+5 \hat{k})+2=0\). Also, find the point of intersection of this line and the plane.

Solution

The required line is perpendicular to the plane

\(\vec{r} \cdot(6 \hat{i}-3 \hat{j}+5 \hat{k})+2=0\) …(1)

So, the required line is parallel to \(\vec{n}=6 \hat{i}-3 \hat{j}+5 \hat{k}\)

Thus, the required line passes through the point with position vector \(\vec{a}=2 \hat{i}-3 \hat{j}-5 \hat{k}\) and is parallel to \(\vec{n}=6 \hat{i}-3 \hat{j}+5 \hat{k}\).

Hence, the vector equation of the required line is

\(\vec{r}=\vec{a}+\lambda \vec{n}, \text { i.e., } \quad \vec{r}=(2 \hat{i}-3 \hat{j}-5 \hat{k})+\lambda(6 \hat{i}-3 \hat{j}+5 \hat{k})\) …(2)

If the line (2) meets the plane (1) then

\([(2 \hat{i}-3 \hat{j}-5 \hat{k})+\lambda(6 \hat{i}-3 \hat{j}+5 \hat{k})] \cdot(6 \hat{i}-3 \hat{j}+5 \hat{k})+2=0\)

⇔ \([(2+6 \lambda) \hat{i}-(3+3 \lambda) \hat{j}+(5 \lambda-5) \hat{k}] \cdot(6 \hat{i}-3 \hat{j}+5 \hat{k})+2=0\)

⇔ 6(2+6λ) + 3(3+3λ) + 5(5λ-5) = -2

⇔ 70λ = 2 ⇔ λ = \(\frac{2}{70}\) = \(\frac{1}{35}\)

Substituting λ = \(\frac{1}{35}\) in (2), we get

\(\vec{r}=(2 \hat{i}-3 \hat{j}-5 \hat{k})+\frac{1}{35}(6 \hat{i}-3 \hat{j}+5 \hat{k})\)

⇔ \(\vec{r}=\frac{1}{35}(76 \hat{i}-108 \hat{j}-170 \hat{k}) .\)

Hence, the required point of intersection is \(\left(\frac{76}{35}, \frac{-108}{35}, \frac{-170}{35}\right)\).

Example 5 Find the angle between the line \(\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{2}\) and the plane 3x + 4y + z + 5 = 0.

Solution

The given line is parallel to \(\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}\).

And, the normal to the given plane is \(\vec{n}=3 \hat{i}+4 \hat{j}+\hat{k} \text {. }\)

Let θ be the angle between the given line and the given plane.

Then, \(\sin \theta=\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}=\frac{(3 \hat{i}-\hat{j}+2 \hat{k}) \cdot(3 \hat{i}+4 \hat{j}+\hat{k})}{|3 \hat{i}-\hat{j}+2 \hat{k}| \cdot|3 \hat{i}+4 \hat{j}+\hat{k}|}\)

= \(\frac{\{3 \times 3+(-1) \times 4+2 \times 1\}}{\left\{\sqrt{3^2+(-1)^2+2^2}\right\}\left\{\sqrt{3^2+4^2+1^2}\right\}}\)

= \(\frac{7}{(\sqrt{14})(\sqrt{26})}=\sqrt{\frac{7}{52}}\)

⇒ \(\theta=\sin ^{-1}\left(\sqrt{\frac{7}{52}}\right) .\)

Hence, the angle between the given line and the given plane is \(\sin ^{-1}\left(\sqrt{\frac{7}{52}}\right) .\)

Example 6 Find the equation of the plane passing through the points (0,0,0) and (3,-1,2) and the parallel to the line \(\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7} \text {. }\)

Solution

Any plane through (0,0,0) is

a(x-0) + b(y-0) + c(z-0) = 0 ⇒ ax + by + cz = 0 …(1)

If this plane passes through (3,-1,2), we have

3a – b + 2c = 0 …(2)

Also, if the plane (1) is parallel to the given line then the normal to this plane is perpendicular to the given line.

∴ a x 1 + b x (-4) + c x 7 = 0 ⇒ a – 4b + 7c = 0 …(3)

Cross multiplying (2) and (3), we have

\(\frac{a}{(-7+8)}=\frac{b}{(2-21)}=\frac{c}{(-12+1)}=k \text { (say) }\)

⇒ \(\frac{a}{1}=\frac{b}{-19}=\frac{c}{-11}=k\)

⇒ a = k, b = -19k and c = -11k.

Substituting a = k, b = -19k and c = -11k in (1), we get the required equation of the plane as

kx – 19ky – 11kz = 0 ⇔ x – 19y – 11z = 0.

Example 7 Find the equation of the plane passing through the points (1,2,3) and (0,-1,0) and parallel to the line \(\frac{x-1}{2}=\frac{y+2}{3}=\frac{z}{-3}\).

Solution

Any plane through (1,2,3) is

a(x-1) + b(y-2) + c(z-3) = 0 …(1)

Since it passes through (0,-1,0), we have

a(0-1) + b(-1-2) + c(0-3) = 0 ⇒ a + 3b + 3c = 0 …(2)

It is being given that the plane (1) is parallel to the line

\(\frac{x-1}{2}=\frac{y+2}{3}=\frac{z}{-3}\)

∴ 2a + 3b 3c = 0 …(3)

On solving (1) and (2), we get

\(\frac{a}{(9+9)}=\frac{b}{(-3-6)}=\frac{c}{(6-3)} \Rightarrow \frac{a}{6}=\frac{b}{-3}=\frac{c}{1}\)

Hence, the required plane is

6(x-1) – 3(y-2) + 1 . (z-3) = 0

⇒ 6x – 3y + z = 3.

Example 8 Find the equation of the plane passing through the point (0,7,-7) and containing the line \(\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\).

Solution

Any plane through (0,7,-7) is

a(x-0) + b(y-7) + c(z+7) = 0 …(1)

If (1) contains the given line then it must pass through the point (-1,3,-2) and must be parallel to the given line.

If (1) passes through (-1,3,-2) then

a(-1-0) + b(3-7) + c(-2+7) = 0 ⇒ a + 4b – 5c = 0 …(2)

If (1) is parallel to the given line then

(-3)a + 2b + 1.c = 0 ⇒ -3a + 2b + c = 0 …(3)

Cross multiplying (2) and (3), we get

\(\frac{a}{(4+10)}=\frac{b}{(15-1)}=\frac{c}{(2+12)}\)

⇒ \(\frac{a}{14}=\frac{b}{14}=\frac{c}{14}\)

⇒ \(\frac{a}{1}=\frac{b}{1}=\frac{c}{1}=k \text { (say) }\)

⇒ a = k, b = k, c = k.

Putting a = k, b = k and c = k in (1), we get the required equation of the plane as

k(x-0) + k(y-7) + k(z+7) = 0

⇒ x + (y-7) + (z+7) = 0 ⇒ x + y + z = 0.

Example 9 Find the equation of the plane passing through the line of intersection of the planes x – 2y + z = 1 and 2x + y + z = 8, and parallel to the line with direction ratios 1,2,1. Find also the perpendicular distance of (1,1,1) from this plane.

Solution

The equation of a plane passing through the intersection of the given planes is

(x – 2y + z – 1) + λ(2x + y + z – 8) = 0

⇔ (1+2λ)x + (λ-2)y + (λ+1)z – (1+8λ) = 0 …(1)

Let this plane be parallel to the line with direction ratios 1, 2, 1.

Then, the normal to this plane is perpendicular to the line having the direction ratios 1, 2, 1.

∴ 1 . (1+2λ) + 2(λ-2) + 1 . (λ+1) = 0 ⇔ 5λ = 2 ⇔ \(\lambda=\frac{2}{5} \text {. }\)

Putting \(\lambda=\frac{2}{5} \text {. }\) in (1), we get the required equation of the plane as 9x – 8y + 7z – 21 = 0.

The length of the perpendicular from (1,1,1) or 9x – 8y + 7z – 21 = 0 is equal to

\(\frac{|9 \times 1-8 \times 1+7 \times 1-21|}{\sqrt{9^2+(-8)^2+7^2}}=\frac{13}{\sqrt{194}} \text { units. }\)

Example 10 Find the equation of the plane passing through the line of intersection of the planes 2x + y – z = 3, 5x – 3y + 4z + 9 = 0, and parallel to the line

\(\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}\)

Solution

The equation of a plane passing through the line of intersection of the given planes is

(2x + y – z – 3) + λ(5x – 3y + 4z + 9) = 0

⇔ (2+5λ)x + (1-3λ)y + (4λ-)λ + (9λ-3) …(1)

If this plane is parallel to the line \(\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}\) then the normal to the plane is perpendicular to this pline.

∴ 2(2+5λ) + 4(1-3λ) + 5(4λ-1) = 0

⇔ 18λ + 3 = 0 ⇔ \(\lambda=\frac{-3}{18}=\frac{-1}{6} .\)

Putting \(\lambda=\frac{-1}{6}\) in (1), we get the required equation of the plane as

\(\left(2-\frac{5}{6}\right) x+\left(1+\frac{3}{6}\right) y+\left(-\frac{4}{6}-1\right) z+\left(-\frac{9}{6}-3\right)=0\)

⇔ 7x + 9y – 10z – 27 = 0.

Example 11 Find the equation of the plane passing through (2,3,-4) and (1,-1,3), and parallel to the x-axis.

Solution

(2,3,-4) and (1,-1,3)

Let the required plane be

a(x-2) + b(y-3) + c(z+4) = 0 …(1)

Since it passes through (1,-1,3), we have

a(1-2) + b(-1-3) + c(3+4) = 0 ⇒ -a -4b + 7c = 0 …(2)

If this plane is parallel to the x-axis then the normal to the plane has the d.r.’s 1,0,0.

∴ a x 1 + b x 0 + c x 0 = 0 ⇒ a = 0.

Putting a = 0 in (2), we get 7c – 4b = 0.

Let b = k. Then, c = \(\frac{4}{7} k .\)

∴ the required plane is

0(x-2)+k(y-3)+\(\frac{4}{7}\) k(z+4)=0

⇒ 7(y-3) + 4(z+4) = 0

⇒ 7y + 4z – 5 = 0.

Hence, the required plane is 7y + 4z – 5 = 0.

Length of Perpendicular from a Point to a Plane

Vector Form

Theorem 1 The length p of the perpendicular drawn from a point with position vector \(\vec{a}\) to the plane \(\vec{r} \cdot \vec{n}=d\) is given by p = \(\frac{|\vec{a} \cdot \vec{n}-d|}{|\vec{n}|} \text {. }\)

Proof

Let π be the given plane whose equation is \(\vec{r} \cdot \vec{n}=d\) …(1)

Then, clearly \(\vec{n}\) is perpendicular to this plane.

Let A be the given point with position vector \(\vec{a}\) and let N be the foot of the perpendicular drawn from A to the given plane. Then , AN = p.

The equation of the line AN, passing through the point \(\vec{a}\) and parallel to \vec{n}, is

\(\vec{r}=\vec{a}+\lambda \vec{n}\), where λ is a scalar …(2)

Clearly, N is the point of intersection of the plane (1) and the line (2).

Substituting \(\vec{r}=\vec{a}+\lambda \vec{n}\) from (2) in (1), we get

\((\vec{a}+\lambda \vec{n}) \cdot \vec{n}=d \Rightarrow \vec{a} \cdot \vec{n}+\lambda(\vec{n} \cdot \vec{n})=d\)

⇒ \(\lambda=\frac{d-(\vec{a} \cdot \vec{n})}{|\vec{n}|^2}\)

Putting this value of λ in (2), we get the position vector of N as

\(\vec{r}=\vec{a}+\left(\frac{d-(\vec{a} \cdot \vec{n})}{|\vec{n}|^2}\right) \vec{n}\)

∴ \(\overrightarrow{A N}\) = (p.v. of N) – (p.v. of A)

= \(\vec{a}+\left(\frac{d-(\vec{a} \cdot \vec{n})}{|\vec{n}|^2}\right) \vec{n}-\vec{a}=\left(\frac{d-(\vec{a} \cdot \vec{n})}{|\vec{n}|^2}\right) \vec{n}\)

⇒ \(A N=|\overrightarrow{A N}|=\left|\frac{\mid d-(\vec{a} \cdot \vec{n})\} \vec{n}}{|\vec{n}|^2}\right|=\frac{|d-(\vec{a} \cdot \vec{n})||\vec{n}|}{|\vec{n}|^2}\)

⇒ \(A N=\frac{|d-(\vec{a} \cdot \vec{n})|}{|\vec{n}|} \Rightarrow p=\frac{|(\vec{a} \cdot \vec{n})-d|}{|\vec{n}|} .\)

Corollary Putting \(\vec{a} = \vec{0}\) in the above formula, we get the length of the perpendicular from the origin to the plane \(\vec{r} \cdot \vec{n}=d\), which is given by

\(p=\frac{|(\overrightarrow{0} \cdot \vec{n})-d|}{|\vec{n}|}=\frac{|-d|}{|\vec{n}|} .\)

Cartesian Form

Theorem 2 The length p of the perpendicular from a point A(x₁,y₁,z₁) to the plane ax + by + cz + d = 0 is given by

\(p=\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right| .\)

Proof

Let N be the foot of the perpendicular drawn from the point A(x₁,y₁,z₁) to the plane ax + by + cz + d = 0.

Then, the equation on this line AN is

\(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=r \text { (say) }\) …(1)

A general point on this line is (x₁ + ar, y₁ + br, z₁ + cr).

This point coincide with N

⇔ it lies on the plane ax + by + cz + d = 0

⇔ a(x₁+ar) + b(y₁+br) + c(z₁+cr) + d = 0

⇔ \(r=\frac{-\left(a x_1+b y_1+c z_1+d\right)}{\left(a^2+b^2+c^2\right)}\) …(2)

∴ AN = \(\left\{\left(x_1+a r-x_1\right)^2+\left(y_1+b r-y_1\right)^2+\left(z_1+c r-z_1\right)^2\right\}^{1 / 2}\)

= \(\sqrt{a^2 r^2+b^2 r^2+c^2 r^2}=\left(\sqrt{a^2+b^2+c^2}\right)|r|\)

= \(\sqrt{a^2+b^2+c^2} \cdot\left|\frac{-\left(a x_1+b y_1+c z_1+d\right)}{\left(a^2+b^2+c^2\right)}\right|\) [using (2)]

= \(\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right| \text {. }\)

Hence, the length p of the perpendicular from A(x1,y1,z1) to the plane ax + by + cz + d = 0 is given by

\(p=\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right| .\)

Corollary Taking x₁ = 0, y₁ = 0 and z₁ = 0, we find that the length of the perpendicular from the origin to the plane ax + by + cz + d = 0 is given by

\(p=\frac{|d|}{\left|\sqrt{a^2+b^2+c^2}\right|} .\)

To Find the Distance between Parallel Planes

Let a₁x + b₁y + c₁z + d₁ = 0 and a₁x + b₁y + c₁z + d₂ = 0 be two parallel planes. Then, we take a point P(x₁,y₁,z₁) on any one of these planes and find the length of the perpendicular drawn from P(x₁,y₁,z₁) to the other plane.

Solved Examples

Example 1 Find the distance of the point \((2 \hat{i}-\hat{j}-4 \hat{k})\) from the plane \(\vec{r} \cdot(3 \hat{i}-4 \hat{j}+12 \hat{k})=9.\)

Solution

We know that the perpendicular distance of a point with position vector \(\vec{a}\) from the plane \(\vec{r} \cdot \vec{n}=d\) is given by

\(p=\frac{|\vec{a} \cdot \vec{n}-d|}{|\vec{n}|} .\)

Here, \(\vec{a}=2 \hat{i}-\hat{j}-4 \hat{k}, \vec{n}=3 \hat{i}-4 \hat{j}+12 \hat{k} \text {, and } d=9 \text {. }\)

∴ the required distance is given by

\(p=\frac{|(2 \hat{i}-\hat{j}-4 \hat{k}) \cdot(3 \hat{i}-4 \hat{j}+12 \hat{k})-9|}{\left|\sqrt{3^2+(-4)^2+(12)^2}\right|}\)

= \(\frac{|(6+4-48)-9|}{|\sqrt{169}|}=\frac{47}{13} \text { units. }\)

Example 2 Find the distance of the point (2,3,4) from the plane \(\vec{r} \cdot(3 \hat{i}-6 \hat{j}+2 \hat{k})+11=0 .\)

Solution

We know that the perpendicular distance of a point with position vector \(\vec{a}\) from the plane \(\vec{r} \cdot \vec{n}=d\) is given by

\(p=\frac{|\vec{a} \cdot \vec{n}-d|}{|\vec{n}|} .\)

Here, \(\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}, \vec{n}=3 \hat{i}-6 \hat{j}+2 \hat{k} \text {, and } d=-11 \text {. }\)

∴ the required distance is given by

\(p=\frac{|(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot(3 \hat{i}-6 \hat{j}+2 \hat{k})+11|}{\left|\sqrt{3^2+(-6)^2+2^2}\right|}\)

= \(\frac{|(6-18+8)+11|}{|\sqrt{49}|}=\frac{7}{7}=1 \text { unit. }\)

Example 3 Find the distance of the point P(2,1,-1) from the plane x – 2y + 4z = 9.

Solution

The required distance

= the length of the perpendicular from P(2,1,-1) to the plane x – 2y + 4z – 9 = 0

= \(\frac{|2-2 \times 1+4 \times(-1)-9|}{\left|\sqrt{1^2+(-2)^2+4^2}\right|}=\frac{13}{\sqrt{21}} \text { units. }\)

Example 4 Find the equation of the planes x + 2y – 2z + 8 = 0, each one of which is at a distance of 2 units from the point (2,1,1).

Solution

Any plane parallel to the plane x + 2y – 2z + 8 = 0 is given by x + 2y – 2z + λ = 0 …(1)

Distance of (1) from (2,1,1) = \(\left|\frac{2+2 \times 1-2 \times 1+\lambda}{\sqrt{1^2+2^2+(-2)^2}}\right|=\left|\frac{2+\lambda}{3}\right|\)

∴ \(\left|\frac{2+\lambda}{3}\right|=2 \Rightarrow \frac{2+\lambda}{3}=2 \quad \text { or } \quad \frac{2+\lambda}{3}=-2\)

⇒ λ = 4 or λ = -8.

Putting these values of λ in (1), we get the required planes as x + 2y – 2z + 4 = 0 and x + 2y – 2z – 8 = 0.

Example 5 Find the distance between the parallel planes 2x – y + 3z + 4 = 0 and 6x – 3y + 9z – 3 = 0.

Solution

2x – y + 3z + 4 = 0 and 6x – 3y + 9z – 3 = 0

Let P(x₁, y₁, z₁) be any point on 2x – y + 3z + 4 = 0.

Then, 2x₁ – y₁ + 3z₁ = -4 …(1)

Length of the perpendicular from P(x₁,y₁,z₁) to 6x – 3y + 9z – 3 = 0 is given by

\(p=\frac{\left|6 x_1-3 y_1+9 z_1-3\right|}{\left|\sqrt{6^2+(-3)^2+9^2}\right|}=\frac{\left|3\left(2 x_1-y_1+3 z_1\right)-3\right|}{|\sqrt{126}|}\)

= \(\frac{|3 \times(-4)-3|}{|\sqrt{126}|}\) [using (1)]

= \(\frac{15}{3 \sqrt{14}}=\frac{5}{\sqrt{14}} \text { units. }\)

Example 6 Find the distance between the parallel planes \(\vec{r} \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})=5 and \vec{r} \cdot(6 \hat{i}-9 \hat{j}+18 \hat{k})+20=0 .\)

Solution

Taking \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\), the equations of the given planes become

\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})=5\)

⇔ 2x – 3y + 6z = 5 …(1)

\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(6 \hat{i}-9 \hat{j}+18 \hat{k})+20=0\)

⇔ 6x – 9y + 18z = -20 …(2)

Let P(x₁,y₁,z₁) be any point on (1).

Then, 2x₁ – 3y₁ + 6z₁ = 5 …(3)

The distance between the given planes

= length of the perpendicular from P(x₁, y₁,z₁) to the plane ()

= \(\frac{\left|6 x_1-9 y_1+18 z_1+20\right|}{\sqrt{6^2+(-9)^2+(18)^2}}=\frac{\left|3\left(2 x_1-3 y_1+6 z_1\right)+20\right|}{\sqrt{441}}\)

= \(\frac{|3 \times 5+20|}{21}\) [using (3)]

= \(\frac{35}{21}=\frac{5}{3} \text { units. }\)

Hence, the distance between the given planes is \frac{5}{3} units.

Example 7 Find the length and the coordinates of the foot of the perpendicular from the point (7,14,5) to the plane 2x + 4y – z = 2.

Solution

Any line through P(7,14,5) and perpendicular to the plane 2x + 4y – z = 2 is given by

\(\frac{x-7}{2}=\frac{y-14}{4}=\frac{z-5}{-1}=\lambda \text { (say) }\)

Any point on this line is

N(2λ + 7, 4λ + 14, -λ + 5).

If N is the foot of the perpendicular from P to the given plane then it must lie on the plane 2x + 4y – z = 2.

∴ 2(2λ + 7) + 4(4λ+14) – (-λ+5) = 2 ⇒ λ = -3.

Thus, we get the point N(1,2,8) on the plane.

Hence, the foot of the perpendicular from P(7,14,5) to the given plane is N(1,2,8).

Length of the perpendicular from P to the given plane

= PN = \(\sqrt{(7-1)^2+(14-2)^2+(5-8)^2}=\sqrt{189}=3 \sqrt{21} \text { units. }\)

Example 8 From the point P(1,2,4), a perpendicular is drawn on the plane 2x + y – 2z + 3 = 0. Find the equation, the length and the coordinates of the foot of the perpendicular.

Solution

Let PN be the perpendicular drawn from the point P(1,2,4) to the plane 2x + y – 2z + 3 = 0.

Then, the equation of the line PN is given by

\(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-4}{-2}=\lambda \text { (say). }\)

So, the coordinates of N are

N(2λ+1, λ+2, -2λ+4).

Since N lies on the plane 2x + y – 2z + 3 = 0, we have

2(2λ+1) + (λ+2) – 2(-2λ+4) + 3 = 0

⇒ \(9 \lambda=1 \Rightarrow \lambda=\frac{1}{9} .\)

∴ Coordinates of N are \(\left(\frac{2}{9}+1, \frac{1}{9}+2, \frac{-2}{9}+4\right) \text {, i.e., }\left(\frac{11}{9}, \frac{19}{9}, \frac{34}{9}\right) \text {. }\)

PN = \(\sqrt{\left(\frac{11}{9}-1\right)^2+\left(\frac{19}{9}-2\right)^2+\left(\frac{34}{9}-4\right)^2}\)

= \(\sqrt{\left(\frac{2}{9}\right)^2+\left(\frac{1}{9}\right)^2+\left(\frac{-2}{9}\right)^2}=\sqrt{\frac{4}{81}+\frac{1}{81}+\frac{4}{81}}=\sqrt{\frac{9}{81}}=\sqrt{\frac{1}{9}}=\frac{1}{3} .\)

Thus, the required equation of PN is \(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-4}{-2}\), coordinates of the foot of the perpendicular are N \(\left(\frac{11}{9}, \frac{19}{9}, \frac{34}{9}\right)\) and length PN = \(\frac{1}{3}\) unit.

Example 9 Find the length and the foot of the perpendicular from the point (1,1,2) to the plane \(\vec{r} \cdot(2 \hat{i}-2 \hat{j}+4 \hat{k})+5=0 .\)

Solution

The given Point is P(1,1,2).

The given plane is

\(\vec{r} \cdot(2 \hat{i}-2 \hat{j}+4 \hat{k})+5=0 .\)

⇔ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}-2 \hat{j}+4 \hat{k})+5=0\)

⇔ 2x – 2y + 4z + 5 = 0 …(1)

Any line through P(1,1,2) and perpendicular to the plane (1) is given by

\(\frac{x-1}{2}=\frac{y-1}{-2}=\frac{z-2}{4}=\lambda \text { (say). }\)

The coordinates of any point N on this line are (2λ+1) – 2(-2λ+1) + 4(4λ+2) + 5 = 0 ⇒ λ = \(-\frac{13}{24}\).

Thus, we get the point N \(\left(\frac{-1}{12}, \frac{25}{12}, \frac{-1}{6}\right) \text {. }\)

Hence, the foot of the perpendicular from P(1,1,2) to the given plane is N \(\left(\frac{-1}{12}, \frac{25}{12}, \frac{-1}{6}\right) \text {. }\)

Length of the perpendicular from P to the given plane

= PN = \(\sqrt{\left(1+\frac{1}{12}\right)^2+\left(1-\frac{25}{12}\right)^2+\left(2+\frac{1}{6}\right)^2}=\frac{13 \sqrt{6}}{12} \text { units. }\)

Example 10 Find the image of the point P(1,3,4) in the plane 2x – y + z + 3 = 0.

Solution

P(1,3,4) in the plane 2x – y + z + 3 = 0

Let Q(x₁,y₁,z₁) be the image of the point P(1,3,4) in the given plane.

The equations of the line through P(1,3,4) and perpendicular to the given plane are \(\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=k \text { (say). }\)

The coordinates of a general point on this line are (2k+1, -k+3, k + 4).

If N is the foot of the perpendicular from P to the given plane then N lies on the plane.

∴ 2(2k+1) – (-k+3) + (k+4) + 3 = 0

⇒ k = -1.

Thus, we get the point N(-1,4,3).

Now,N is the midpoint of PQ.

∴ \(\frac{1+x_1}{2}=-1, \frac{3+y_1}{2}=4, \frac{4+z_1}{2}=3\)

⇒ x₁ = -3, y₁ – 5, z₁ = 2.

Hence, the required image of P(1,3,4) is Q(-3,5,2).

Example 11 A variable plane which remains at a constant distance 3p from the origin cuts the coordinates axes at A, B, C. Show that the locus of the centroid of △ABC is x^-2 + y^-2 + z^-2 = p^-2.

Solution

Let the equation of the variable plane be

\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) …(1)

This plane meets the x-axis, y-axis and z-axis at the points A(a,0,0), B(0,b,0) and C(0,0,c) respectively. Let (α, β, γ) be the coordinates of the centroid of △ABC.

Then, \(\alpha=\frac{a+0+0}{3}, \beta=\frac{0+b+0}{3} \text { and } \gamma=\frac{0+0+c}{3}\)

⇒ \(\alpha=\frac{a}{3}, \beta=\frac{b}{3} \text { and } \gamma=\frac{c}{3}\)

⇒ a = 3α, b = 3β, c = 3γ …(3)

∴ 3p = length of the perpendicular from (0,0,0) to the plane (1)

⇒ \(3 p=\frac{\left|\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1\right|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\)

⇒ \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\frac{1}{3 p}\)

⇒ \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{9 p^2}\)

⇒ \(\frac{1}{9 \alpha^2}+\frac{1}{9 \beta^2}+\frac{1}{9 \gamma^2}=\frac{1}{9 p^2}\) [using (2)]

⇒ α^-2 + β^-2 + γ^-2 = p^-2.

Hence, the required locus is x^-2 + y^-2 + z^-2 = p^-2.

Example 12 A variable plane is at a constant distance p from the origin and meets the coordinate axes in A, B, C. Show that the locus of the centroid of the tetrahedron OABC is x^-2 + y^-2 + z^-2 = 16p^-2.

Solution

Let the equation of the variable plane be

\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) …(1)

This plane meets the x-axis, y-axis and z-axis at the points A(a,0,0), B(0,b,0) and C(0,0,c) respectively.

Let (α, β, γ) be the coordinates of the centroid of the tetrahedron OABC.

Then, \(\alpha=\frac{0+a+0+0}{4}=\frac{a}{4}, \beta=\frac{0+0+b+0}{4}=\frac{b}{4}\)

and \(\gamma=\frac{0+0+0+c}{4}=\frac{c}{4} \Rightarrow a=4 \alpha, b=4 \beta, c=4 \gamma\) …(2)

∴ p = distance of the plane (1) from (0,0,0)

⇒ \(p=\frac{\left|\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1\right|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\)

⇒ \(\frac{1}{p}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\frac{1}{p^2}=\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

⇒ \(\frac{1}{p^2}=\frac{1}{16 \alpha^2}+\frac{1}{16 \beta^2}+\frac{1}{16 \gamma^2}\) [using (2)]

⇒ \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=\frac{16}{p^2} \Rightarrow \alpha^{-2}+\beta^{-2}+\gamma^{-2}=16 p^{-2} .\)

Hence, the required locus is x^-2 + y^-2 + z^-2 = 16p^-2.

Example 13 Find the distance of the point (2,3,4) from the plane 3x + 2y + 2z + 5 = 0, measured parallel to the line \(\frac{x+3}{3}=\frac{y-2}{6}=\frac{z}{2} .\)

Solution

The point (2,3,4) from the plane 3x + 2y + 2z + 5 = 0

Let l be the given line \(\frac{x+3}{3}=\frac{y-2}{6}=\frac{z}{2}\), and let P(2,3,4) be the given point.

Let PQ || l.

Then, PQ is the line passing through P(2,3,4) and having direction ratios 3,6,2.

So, the equations of PQ are

\(\frac{x-2}{3}=\frac{y-3}{6}=\frac{z-4}{2}=\lambda \text { (say). }\)

The coordinates of any point Q on this line are (3λ+2, 6λ+3, 2λ+4).

If this point Q lies on the given plane then

3(3λ+2) + 2(6λ+3) + 2(2λ+4) + 5 = 0

⇔ 25λ + 25 = 0 ⇔ 25λ = -25 ⇔ λ = -1.

So, the coordinates of Q are (-1,-3,2).

∴ the requried distance = PQ = \(\sqrt{(2+1)^2+(3+3)^2+(4-2)^2}\)

= √49 = 7 units.

Example 14 Find the distance of the point (-2,3,-4) from the line \(\frac{x+2}{3}=\frac{2 y+3}{4}=\frac{3 z+4}{5}\), measured parallel to the plane 4x + 12y – 3z + 1 = 0.

Solution

\(\frac{x+2}{3}=\frac{2 y+3}{4}=\frac{3 z+4}{5}\)

Let P(-2,3,-4) be the given point.

Let \(\frac{x+2}{3}=\frac{2 y+3}{4}=\frac{3 z+4}{5}=\lambda\) be the given line.

A general point on this line is \(Q\left(3 \lambda-2, \frac{4 \lambda-3}{2}, \frac{5 \lambda-4}{3}\right)\)

Direction ratios of PQ are

\((3 \lambda-2+2),\left(\frac{4 \lambda-3}{2}-3\right),\left(\frac{5 \lambda-4}{3}+4\right), \text { i.e. } 3 \lambda, \frac{4 \lambda-9}{2}, \frac{5 \lambda+8}{3} \text {. }\)

Direction ratios of normal to the plane are 4,12,-3.

PQ is parallel to the given plane

⇒ PQ is perpendicular to the normal to the plane

⇒ \((4 \times 3 \lambda)+12 \times \frac{(4 \lambda-9)}{2}-3 \times \frac{(5 \lambda+8)}{3}=0\)

⇒ 12λ + 6(4λ-9) – (5λ+8) = 0 ⇒ 31λ = 62 ⇒ λ = 2.

∴ coordinates of Q are \(\left(4, \frac{5}{2}, 2\right)\).

∴ PQ = \(\sqrt{(4+2)^2+\left(\frac{5}{2}-3\right)^2+(2+4)^2}=\sqrt{36+\frac{1}{4}+36}\)

= \(\sqrt{\frac{289}{4}}=\frac{17}{2} \text { units. }\)

Condition for the Coplanarity of Two Lines

Vector Form

Theorem 1 The condition for two lines \(\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1} \text { and } \vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\) to be coplanar is that

\(\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=0 \text {, i.e., }\left[\overrightarrow{a_2}-\overrightarrow{a_1} \overrightarrow{b_1} \overrightarrow{b_2}\right]=0 \text {. }\)

Also, the equation of the plane containing both these lines is

\(\left(\vec{r}-\vec{a}_1\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=0 \quad \text { or } \quad\left(\vec{r}-\vec{a}_2\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)=0 \text {. }\)

Proof

The equations of the given lines are

\(\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}\) …(1)

\(\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\) …(2)

The line (1) passes through a point A with position vector \(\overrightarrow{a_1}\) and is parallel to \(\overrightarrow{b_1}\).

The line (2) passes through a point B with position vector \(\overrightarrow{a_2}\) and is parallel to \(\overrightarrow{b_2}\).

Now, \(\overrightarrow{A B}=(\text { p.v. of } B)-(\text { p.v. of } A)=\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \text {. }\)

∴ the given lines are coplanar

⇔ \(\overrightarrow{A B}, \overrightarrow{b_1}, \overrightarrow{b_2}\) are coplanar

⇔ \(\overrightarrow{A B} \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=0\)

⇔ \(\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=0 .\)

Equation of the Plane Containing Both the Lines

If the lines \(\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1} \text { and } \vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\) are coplanar then their common plane is perpendicular to the vector \(\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)\), and this plane passes through each of the points \(\overrightarrow{a_1} and \overrightarrow{a_2}\).

So, its equation is

\(\left(\vec{r}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=0 \text {, or }\left(\vec{r}-\overrightarrow{a_2}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=0 \text {. }\)

Cartesian Form

Theorem 2 The lines \(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} \text {, and } \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\)

are coplanar ⇔ \(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=0 .\)

And, the equation of the plane containing both these lines is

\(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=0 \text { or }\left|\begin{array}{ccc}
x-x_2 & y-y_2 & z-z_2 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=0 \text {. }\)

Proof

The given lines are

\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\) …(1)

\(\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\) …(2)

The line (1) passes through a point A(x₁,y₁,z₁) and is parallel to the vector \(\overrightarrow{u_1}=a_1 \hat{i}+b_1 \hat{j}+c_1 \hat{k}.\)

Also, the line (2) passes through a point B(x₂,y₂,z₂) and is parallel to the vector \(\vec{u}_2=a_2 \hat{i}+b_2 \hat{j}+c_2 \hat{k} .\)

Now, \(\overrightarrow{A B}=(\text { p.v. of } B)-(\text { p.v, of } A)\)

= \(\left(x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\right)-\left(x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\right)\)

= \(\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k} .\)

∴ lines (1) and (2) are coplanar

⇔ there is a plane which passes through the points A and B, and which is parallel to each of \(\overrightarrow{u_1} \text { and } \overrightarrow{u_2}\)

⇔ \(\overrightarrow{A B}, \overrightarrow{u_1}, \overrightarrow{u_2}\) are parallel to the same plane

⇔ \(\overrightarrow{A B}, \overrightarrow{u_1}, \overrightarrow{u_2}\) are coplanar

⇔ \(\left[\begin{array}{lll}
\overrightarrow{A B} & \overrightarrow{u_1} & \overrightarrow{u_2}
\end{array}\right]=0\)

⇔ \(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=0 .\)

Equation of the Plane Containing Both the Lines

If the lines (1) and (2) are coplanar then their common plane is the plane containing the line (1) and parallel to the line (2), or it is the plane containing the line (2) and parallel to the line (1).

Hence, its equation is

\(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=0 \text {, or }\left|\begin{array}{ccc}
x-x_2 & y-y_2 & z-z_2 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=0 \text {. }\)

Solved Examples

Example 1 Show that the lines

\(\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3 \hat{i}-\hat{j}) \text { and } \vec{r}=(4 \hat{i}-\hat{k})+\mu(2 \hat{i}+3 \hat{k})\)

are coplanar. Also, find the equation of the plane containing both these lines.

Solution

We know that the lines \(\vec{r}=\vec{a}_1+\lambda \overrightarrow{b_1} \text { and } \vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\) are coplanar

⇔ \(\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=0 .\)

Here, \(\overrightarrow{a_1}=\hat{i}+\hat{j}-\hat{k}, \overrightarrow{b_1}=3 \hat{i}-\hat{j}, \overrightarrow{a_2}=4 \hat{i}-\hat{k} \text { and } \vec{b}_2=2 \hat{i}+3 \hat{k} \text {. }\)

∴ \(\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)=(4 \hat{i}-\hat{k})-(\hat{i}+\hat{j}-\hat{k})=(3 \hat{i}-\hat{j}) .\)

\(\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
3 & -1 & 0 \\
2 & 0 & 3
\end{array}\right|\)

= \((-3-0) \hat{i}-(9-0) \hat{j}+(0+2) \hat{k}=-3 \hat{i}-9 \hat{j}+2 \hat{k}\)

∴ \(\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=(3 \hat{i}-\hat{j}) \cdot(-3 \hat{i}-9 \hat{j}+2 \hat{k})\)

= [3 x (-3) + (1) x (-9) + 0 x 2] = 0.

Hence, the given lines are coplanar.

The equation of the plane containing both the given lines is given by

\(\left(\vec{r}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)=0\)

⇔ \(\{\vec{r}-(\hat{i}+\hat{j}-\hat{k})\} \cdot(-3 \hat{i}-9 \hat{j}+2 \hat{k})=0\)

⇔ \(\vec{r} \cdot(-3 \hat{i}-9 \hat{j}+2 \hat{k})=(\hat{i}+\hat{j}-\hat{k}) \cdot(-3 \hat{i}-9 \hat{j}+2 \hat{k})\)

⇔ \(\vec{r} \cdot(-3 \hat{i}-9 \hat{j}+2 \hat{k})=(-3-9-2)\)

⇔ \(\vec{r} \cdot(3 \hat{i}+9 \hat{j}-2 \hat{k})+14=0\)

The equation of the plane containing both these lines \(\vec{r} \cdot(3 \hat{i}+9 \hat{j}-2 \hat{k})+14=0\)

Example 2 Show that the lines \(\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3} \text { and } \frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}\) are coplanar. Also, find the equation of the plane containing these lines.

Solution

We know that the lines

\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} \text {, and } \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\)

are coplanar ⇔ \(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=0 \text {, }\)

and the equation of the plane containing these lines is

\(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=0 .\)

Here, x₁ = 0, y₁ = 2, z₁ = -3; x₂ = 2, y₂ = 6, z₂ = 3; a₁ = 1, b₁ = 2, c₁ = 3; a₂ = 2, b₂ = 3, c₂ = 4.

∴ \(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=\left|\begin{array}{ccc}
2 & 4 & 6 \\
1 & 2 & 3 \\
2 & 3 & 4
\end{array}\right|=0 .\)

Hence, the two given lines are coplanar.

The equation of the plane containing both these lines is

\(\left|\begin{array}{ccc}
x-0 & y-2 & z+3 \\
1 & 2 & 3 \\
2 & 3 & 4
\end{array}\right|=0 \Leftrightarrow\left|\begin{array}{ccc}
x & y-2 & z+3 \\
1 & 2 & 3 \\
2 & 3 & 4
\end{array}\right|=0\)

⇔ x(8-9) – (y-2)(4-6) + (z+3)(3-4) = 0

⇔ -x + 2(y-2) – (z+3) = 0 ⇔ x – 2y + z+ 7 = 0.

Hence, the required plane is x – 2y + z + 7 = 0.

Example 3 Show that the lines

\(\frac{x-a+d}{\alpha-\delta}=\frac{y-a}{\alpha}=\frac{z-a-d}{\alpha+\delta} \text {, and } \frac{x-b+c}{\beta-\gamma}=\frac{y-b}{\beta}=\frac{z-b-c}{\beta+\gamma}\)

are coplanar.

Solution

The given lines are

\(\frac{x-(a-d)}{\alpha-\delta}=\frac{y-a}{\alpha}=\frac{z-(a+d)}{\alpha+\delta}\) …(1)

\(\frac{x-(b-c)}{\beta-\gamma}=\frac{y-b}{\beta}=\frac{z-(b+c)}{\beta+\gamma}\) …(2)

We know that the lines

\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} \text {, and } \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\)

are coplanar ⇔ \(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=0 .\)

Here, x₁ = (a-d), y₁ = a, z₁ = (a+d); x₂ = (b-c), y₂ = b, z₂ = (b+c); a₁ = (α-δ), b₁ = α, c₁ = (α+δ); and a₂ = (β-γ), b₂ = β, c₂ = (β+γ).

∴ \(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
(b-c)-(a-d) & b-a & (b+c)-(a+d) \\
\alpha-\delta & \alpha & \alpha+\delta \\
\beta-\gamma & \beta & \beta+\gamma
\end{array}\right|\)

= \(\left|\begin{array}{ccc}
2(b-a) & b-a & b+c-a+d \\
2 \alpha & \alpha & \alpha+\delta \\
2 \beta & \beta & \beta+\gamma
\end{array}\right| \quad\left\{C_1 \rightarrow C_1+C_3\right\}\)

= 0 [∵ C₁ and C₂ are proportional].

Hence, the given lines are coplanar.

Example 4 Find the equation of the plane which contains the two parallel lines \(\frac{x-3}{3}=\frac{y+4}{2}=\frac{z-1}{1} \text {, and } \frac{x+1}{3}=\frac{y-2}{2}=\frac{z}{1} \text {. }\)

Solution

\(\frac{x-3}{3}=\frac{y+4}{2}=\frac{z-1}{1} \text {, and } \frac{x+1}{3}=\frac{y-2}{2}=\frac{z}{1} \text {. }\)

The plane which contains the two given parallel lines must pass through the points (3,-4,1) and (-1,2,0), and must be parallel to the line having direction ratios 3,2,1.

Any plane passing through (3,-4,1) is

a(x-3) + b(y+4) + c(z-1) = 0 …(1)

If this plane passes through the point (-1,2,0) then

a(-1-3) + b(2+4) + c(0-1) = 0 ⇒ 4a – 6b + c = 0 …(2)

If the plane (1) is parallel to the line having direction ratios 3,2,1 then

3a + 2b + c = 0 …(3)

Cross multiplying (2) and (3), we get

\(\frac{a}{(-6-2)}=\frac{b}{(3-4)}=\frac{c}{(8+18)} \Leftrightarrow \frac{a}{-8}=\frac{b}{-1}=\frac{c}{26}\)

⇔ \(\frac{a}{8}=\frac{b}{1}=\frac{c}{-26}=k \text { (say) }\)

⇔ a = 8k, b = k, c = -26k.

Putting these values of a,b,c in (1), we get the required equation of the plane as

8k(x-3) + k(y+4) – 26k(z-1) = 0

⇔ 8(x-3) + (y+4) – 26(z-1) = 0 ⇔ 8x + y – 26z + 6 = 0.

The equation of the plane which contains the two parallel lines 8x + y – 26z + 6 = 0.