# NCERT Solutions For Class 12 Maths Chapter 11 Exercise 11.3

Go back to  'Three Dimensional Geometry'

## Chapter 11 Ex.11.3 Question 1

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

(a) $$z = 2$$

(b) $$x + y + z = 1$$

(c) $${\rm{2x + 3y - z}} = 5$$

(d) $$5y + 8 = 0$$

### Solution

(a) The equation of the plane is $$z = 2$$ or $$0x + 0y + z = 2 \quad \ldots \left( 1 \right)$$

The direction ratios of normal are $$0,0$$ and $$1$$.

Therefore,

$\sqrt {{0^2} + {0^2} + {1^2}} = 1$

Dividing both sides of equation (1) by $$1,$$ we obtain

$0.x + 0.y + 1.z = 2$

This is of the form $$lx + my + nz = d$$, where $$l,\;m,\;n$$ are the direction cosines of normal to the plane and $$d$$ is the distance of the perpendicular drawn from the origin.

Hence, the direction cosines are $$0,0$$ and $$1$$ and the distance of the plane form the origin is $$2$$ units.

(b) $$x + y + z = 1 \quad \ldots \left( 1 \right)$$

The direction ratios of normal are $$1,1$$ and $$1$$.

Therefore,

$\sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt 3$

Dividing both sides of equation (1) by $$\sqrt 3$$ , we get

$\frac{1}{{\sqrt 3 }}x + \frac{1}{{\sqrt 3 }}y + \frac{1}{{\sqrt 3 }}z = \frac{1}{{\sqrt 3 }}$

This equation is one of the form $$lx + my + nz = d$$, where $$l,\;m,\;n$$ are direction cosines of normal to the plane and $$d$$ is the distance of normal from the origin.

Hence, the direction cosines of the normal are $$\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}$$ and $$\frac{1}{{\sqrt 3 }}$$ and the distance of normal form the origin is $$\frac{1}{{\sqrt 3 }}$$ units.

(c) $$2x + 3y - z = 5 \quad \ldots \left( 1 \right)$$

The direction ratios of normal are $$2,3$$ and $$- 1$$.

Therefore,

$\sqrt {{{\left( 2 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( { - 1} \right)}^2}} = \sqrt {14}$

Dividing both sides of equation (1) by $$\sqrt {14}$$ , we get

$\frac{2}{{\sqrt {14} }}x + \frac{3}{{\sqrt {14} }}y - \frac{1}{{\sqrt {14} }}z = \frac{5}{{\sqrt {14} }}$

This equation is one of the form $$lx + my + nz = d$$, where $$l,\;m,\;n$$ are direction cosines of normal to the plane and $$d$$ is the distance of normal from the origin.

Hence, the direction cosines of the normal are $$\frac{2}{{\sqrt {14} }},\frac{3}{{\sqrt {14} }}$$ and $$\frac{{ - 1}}{{\sqrt {14} }}$$ and the distance of normal form the origin is $$\frac{5}{{\sqrt {14} }}$$ units.

(d) $$5y + 8 = 0$$

$\Rightarrow \; 0x + 5y + 0z = 8 \quad \ldots \left( 1 \right)$

The direction ratios of normal are $$0, - 5$$ and $$0$$.

Therefore,

$\sqrt {{0^2} + {{\left( { - 5} \right)}^2} + {0^2}} = 5$

Dividing both sides of equation (1) by $$5,$$ we get

$0x + y + 0z = \frac{8}{5}$

This equation is one of the form $$lx + my + nz = d$$, where $$l,\;m,\;n$$ are direction cosines of normal to the plane and $$d$$ is the distance of normal from the origin.

Hence, the direction cosines of the normal to the plane are $$0,1$$ and $$0$$ and the distance of normal form the origin is $$\frac{8}{5}$$ units.

## Chapter 11 Ex.11.3 Question 2

Find the vector equation of a plane which is at the distance of $$7$$ units from the origin and normal to the vector $$3\hat i + 5\hat j - 6\hat k$$.

### Solution

The normal vector is, $$\vec n = 3\hat i + 5\hat j - 6\hat k$$

\begin{align}\hat n &= \frac{{\vec n}}{{\left| {\vec n} \right|}} = \frac{{3\hat i + 5\hat j - 6\hat k}}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( 5 \right)}^2} + {{\left( 6 \right)}^2}} }}\\&= \frac{{3\hat i + 5\hat j - 6\hat k}}{{\sqrt {70} }}\end{align}

The equation of the plane with position vector $$\vec r$$ is given by, $$\vec r.\hat n = d$$

Hence,

$\vec r.\left( {\frac{{3\hat i + 5\hat j - 6\hat k}}{{\sqrt {70} }}} \right) = 7$

## Chapter 11 Ex.11.3 Question 3

Find the Cartesian equation of the following planes:

(a) $$\vec r.\left( {\hat i + \hat j - \hat k} \right) = 2$$

(b) $$\vec r.\left( {2\hat i + 3\hat j - 4\hat k} \right) = 1$$

(c) $$\vec r.\left[ {\left( {s - 2t} \right)\hat i + \left( {3 - t} \right)\hat j + \left( {2s + t} \right)\hat k} \right] = 15$$

### Solution

(a) Given equation of the plane is

$\vec r.\left( {\hat i + \hat j - \hat k} \right) = 2 \qquad \ldots \left( 1 \right)$

For any arbitrary point, $$P\left( {x,y,z} \right)$$ on the plane, position vector $$\vec r$$ is given by, $$\vec r = x\hat i + y\hat j - z\hat k$$

Putting the values of $$\vec r$$ in equation $$\left( 1 \right)$$, we get

\begin{align}&\left( {x\hat i + y\hat j - z\hat k} \right).\left( {\hat i + \hat j - \hat k} \right) = 2\\& \Rightarrow \; x + y - z = 2\end{align}

(b) $$\vec r.\left( {2\hat i + 3\hat j - 4\hat k} \right) = 1 \qquad \ldots \left( 1 \right)$$

For any arbitrary point $$P\left( {x,y,z} \right)$$ on the plane, position vector $$\vec r$$ is given by, $$\vec r = x\hat i + y\hat j - z\hat k$$

Putting the values of $$\overrightarrow r$$ in equation $$\left( 1 \right)$$, we get

\begin{align}&\left( {x\hat i + y\hat j - z\hat k} \right).\left( {2\hat i + 3\hat j - 4\hat k} \right) = 1\\& \Rightarrow 2x + 3y - 4z = 1\end{align}

(c) $$\vec r.\left[ {\left( {s - 2t} \right)\hat i + \left( {3 - t} \right)\hat j + \left( {2s + t} \right)\hat k} \right] = 15 \quad \ldots \left( 1 \right)$$

For any arbitrary point, $$P\left( {x,y,z} \right)$$ on the plane, position vector $$\vec r$$ is given by,

$$\vec r = x\hat i + y\hat j - z\hat k$$

Putting the values of $$\vec r$$ in equation $$\left( 1 \right)$$, we get

\begin{align}&\left( {x\hat i + y\hat j - z\hat k} \right).\left[ {\left( {s - 2t} \right)\hat i + \left( {3 - t} \right)\hat j + \left( {2s + t} \right)\hat k} \right] = 15\\& \Rightarrow\; \left( {s - 2t} \right)x + \left( {3 - t} \right)y + \left( {2s + t} \right)z = 15\end{align}

## Chapter 11 Ex.11.3 Question 4

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

(a) $$2x + 3y + 4z-12 = 0$$

(b) $$3y + 4z-6 = 0$$

(c) $$x + y + z = 1$$

(d) $$5y + 8 = 0$$

### Solution

(a) Let the coordinates of the foot of perpendicular $$P$$ from the origin to the plane be $$\left( {{x_1},{y_1},{z_1}} \right)$$

$2x + 3y + 4z--12 = 0 \qquad \ldots \left( 1 \right)$

The direction ratios of normal are $$2,3$$ and $${\rm{4}}$$

Therefore,

$\sqrt {{{\left( 2 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 4 \right)}^2}} = \sqrt {29}$

Dividing both sides of equation (1) by $$\sqrt {29}$$, we get

$\frac{2}{{\sqrt {29} }}x + \frac{3}{{\sqrt {29} }}y + \frac{4}{{\sqrt {29} }}z = \frac{{12}}{{\sqrt {29} }} \qquad \ldots \left( 2 \right)$

This equation is one of the form $$lx + my + nz = d$$, where $$l,\;m,\;n$$ are direction cosines of normal to the plane and $$d$$ is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by $$\left( {ld,md,nd} \right)$$

Hence, the coordinates of the foot of the perpendicular are

\begin{align}&\left( {\frac{2}{{\sqrt {29} }} \times \frac{{12}}{{\sqrt {29} }},\frac{3}{{\sqrt {29} }} \times \frac{{12}}{{\sqrt {29} }},\frac{4}{{\sqrt {29} }} \times \frac{{12}}{{\sqrt {29} }}} \right)\\& \Rightarrow \;\left( {\frac{{24}}{{29}},\frac{{36}}{{29}},\frac{{48}}{{29}}} \right)\end{align}

(b) Let the coordinates of the foot of perpendicular $$P$$ from the origin to the plane be$$\left( {{x_1},{y_1},{z_1}} \right)$$

$3y + 4z--6 = 0\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$

The direction ratios of the normal are $$0,3$$ and $$4$$.

Therefore,

$\sqrt {{0^2} + {3^2} + {4^2}} = 5$

Dividing both sides of equation (1) by $$5,$$ we get

$0x + \frac{3}{5}y + \frac{4}{5}z = \frac{6}{5}$

This equation is one of the form $$lx + my + nz = d$$, where $$l,\;m,\;n$$ are direction cosines of normal to the plane and $$d$$ is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by $$\left( {ld,md,nd} \right)$$

Hence, the coordinates of the foot of the perpendicular are

\begin{align}&\left( {0 \times \frac{6}{5},\frac{3}{5} \times \frac{6}{5},\frac{4}{5} \times \frac{6}{5}} \right)\\& \Rightarrow \ \left( {0,\frac{{18}}{{25}},\frac{{24}}{{25}}} \right)\end{align}

(c) Let the coordinates of the foot of perpendicular $$P$$ from the origin to the plane be

$x + y + z = 1 \qquad \ldots \left( 1 \right)$

The direction ratios of the normal are $$1,1$$ and $$1$$.

Therefore,

$\sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt 3$

Dividing both sides of equation (1) by $$\sqrt 3$$, we get,

$\frac{1}{{\sqrt 3 }}x + \frac{1}{{\sqrt 3 }}y + \frac{1}{{\sqrt 3 }}z = \frac{1}{{\sqrt 3 }}$

This equation is one of the form $$lx + my + nz = d$$, where $$l,\;m,\;n$$ are direction cosines of normal to the plane and $$d$$ is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by $$\left( {ld,md,nd} \right)$$

Hence, the coordinates of the foot of the perpendicular are

\begin{align}&\left( {0 \times \frac{6}{5},\frac{3}{5} \times \frac{6}{5},\frac{4}{5} \times \frac{6}{5}} \right)\\& \Rightarrow\; \left( {0,\frac{{18}}{{25}},\frac{{24}}{{25}}} \right)\end{align}

(d) Let the coordinates of the foot of perpendicular $$P$$ from the origin to the plane be

\begin{align}&5y + 8 = 0\\&\Rightarrow\;0x--5y + 0z = 8 \qquad \ldots \left( 1 \right) \end{align}

The direction ratios of the normal are $$0, - 5$$ and $$0$$.

Therefore,

$\sqrt {{0^2} + {{\left( { - 5} \right)}^2} + 0} = 5$

Dividing both sides of equation (1) by $$5,$$ we obtain

$0x--y + 0z = \frac{8}{5}$

This equation is one of the form $$lx + my + nz = d$$, where $$l,\;m,\;n$$ are direction cosines of normal to the plane and $$d$$ is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by $$\left( {ld,md,nd} \right)$$

Hence, the coordinates of the foot of the perpendicular are

\begin{align}&\left( {0 \times \frac{8}{5}, - 1 \times \frac{8}{5},0 \times \frac{8}{5}} \right)\\&\Rightarrow \;\left( {0, - \frac{8}{5},0} \right)\end{align}

## Chapter 11 Ex.11.3 Question 5

Find the vector and Cartesian equation of the planes

(a) that passes through the point $$\left( {{\rm{1}},0, - {\rm{2}}} \right)$$ and the normal to the plane is $$\hat i + \hat j - \hat k$$

(b) that passes through the point $$\left( {{\rm{1}},{\rm{4}},6} \right)$$and the normal vector to the plane is .

### Solution

(a) The position vector of point $$\left( {{\rm{1}},0, - {\rm{2}}} \right)$$ is $$\vec a = \hat i - 2\hat k$$

The normal vector $$\vec N$$ perpendicular to the plane is $$\vec N = \hat i + \hat j - \hat k$$

The vector equation of the plane is given by,

\begin{align}& \Rightarrow \left( {\vec r - \vec a} \right).\vec N = 0\\ &\Rightarrow \left[ {\vec r - \left( {\vec i - 2\hat k} \right)} \right].\left( {\hat i + \hat j - \hat k} \right) = 0\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Since, $$\vec r$$ is the positive vector of any point $$P\left( {x,y,z} \right)$$in the plane.

Hence,

$$\vec r = x\hat i + y\hat j + z\hat k$$

Thus, equation (1) becomes

\begin{align} &\Rightarrow \left[ {\left( {x\hat i + y\hat j + z\hat k} \right) - \left( {\hat i - 2\hat k} \right)} \right].\left( {\hat i + \hat j - \hat k} \right) = 0\\ &\Rightarrow \left[ {\left( {x - 1} \right)\hat i + y\hat j + \left( {z + 2} \right)\hat k} \right].\left( {\hat i + \hat j - \hat k} \right) = 0\\&\Rightarrow \left( {x - 1} \right) + y - \left( {z + 2} \right) = 0\\ &\Rightarrow x + y - z - 3 = 0\\ &\Rightarrow x + y - z = 3\end{align}

(b) The position vector of point $$\left( {{\rm{1}},{\rm{4}},6} \right)$$ is $$\vec a = \hat i + 4\hat j + 6\hat k$$

The normal vector $$\vec N$$ perpendicular to the plane is $$\vec N = \hat i - 2\hat j + \hat k$$

The vector equation of the plane is given by,

\begin{align} &\Rightarrow \left( {\vec r - \vec a} \right).\vec N = 0\\ &\Rightarrow \left[ {\vec r - \left( {\hat i + 4\hat j + 6\hat k} \right)} \right].\left( {\hat i - 2\hat j + \hat k} \right) = 0\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

Since, $$\vec r$$ is the positive vector of any point $$P\left( {x,y,z} \right)$$in the plane.

Hence,

$$\vec r = x\hat i + y\hat j + z\hat k$$

Thus, equation (1) becomes

\begin{align}& \Rightarrow \left[ {\left( {x\hat i + y\hat j + z\hat k} \right) - \left( {\hat i + 4\hat j + 6\hat k} \right)} \right].\left( {\hat i - 2\hat j + \hat k} \right) = 0\\ &\Rightarrow \left[ {\left( {x - 1} \right)\hat i + \left( {y - 4} \right)\hat j + \left( {z - 6} \right)\hat k} \right].\left( {\hat i - 2\hat j + \hat k} \right) = 0\\& \Rightarrow \left( {x - 1} \right) - 2\left( {y - 4} \right) + \left( {z - 6} \right) = 0\\& \Rightarrow x - 2y + z + 1 = 0\end{align}

## Chapter 11 Ex.11.3 Question 6

Find the equations of the planes that passes through the points.

(a) $$\left( {1,1, - 1} \right),\left( {6,4, - 5} \right),\left( { - 4,2,3} \right)$$

(b) $$\left( {1,1,0} \right),\left( {1,2,1} \right),\left( { - 2,2, - 1} \right)$$

### Solution

(a) The given points are $$A\left( {1,1, - 1} \right),B\left( {6,4, - 5} \right)$$ and$$C\left( { - 4,2,3} \right)$$.

\begin{align}\left( {\begin{array}{*{20}{c}}1&1&{ - 1}\\6&4&{ - 5}\\{ - 4}&{ - 2}&3\end{array}} \right) &= \left( {12 - 10} \right) - \left( {18 - 20} \right) - \left( { - 12 + 16} \right)\\ &= 2 + 2 - 4\\& = 0\end{align}

Since, the points are collinear, there will be infinite number of planes passing through the given points.

(b) The given points are $$A\left( {1,1,0} \right),B\left( {1,2,1} \right)$$ and $$C\left( { - 2,2, - 1} \right)$$.

\begin{align}\left( {\begin{array}{*{20}{c}}1&1&0\\1&2&1\\{ - 2}&2&{ - 1}\end{array}} \right) &= \left( { - 2 - 2} \right) - \left( {2 + 2} \right)\\ &= - 8\\ &\ne 0\end{align}

Thus, a plane will pass through the points.

The equation of the plane through the points $$\left( {{x_1},{y_1},{z_1}} \right),\left( {{x_2},{y_2},{z_2}} \right)$$ and $$\left( {{x_3},{y_3},{z_3}} \right)$$is given by

\begin{align}& \Rightarrow \left( {\begin{array}{*{20}{c}}{x - {x_1}}&{y - {y_1}}&{z - {z_1}}\\{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\{{x_3} - {x_1}}&{{y_3} - {y_1}}&{{z_3} - {z_1}}\end{array}} \right) = 0\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{x - 1}&{y - 1}&z\\0&1&1\\{ - 3}&1&{ - 1}\end{array}} \right) = 0\\ &\Rightarrow \left( { - 2} \right)\left( {x - 1} \right) - 3\left( {y - 1} \right) + 3z = 0\\& \Rightarrow - 2x + 2 - 3y + 3 + 3z = 0\\& \Rightarrow - 2x - 3y + 3z + 5 = 0\\ &\Rightarrow 2x + 3y - 3z = 5\end{align}

## Chapter 11 Ex.11.3 Question 7

Find the intercepts cut off by the plane $$2x + y - z = 5$$

### Solution

$$2x + y - z = 5$$

Dividing both sides of equation by $$5$$, we get

\begin{align} \Rightarrow \frac{2}{5}x + \frac{y}{5} - \frac{z}{5} = 1\\ \Rightarrow \frac{x}{{\frac{5}{2}}} + \frac{y}{5} + \frac{z}{{ - 5}} = 1\end{align}

The equation of a plane in intercept form is $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$ , where $$a,b$$ and $$c$$ are intercepts cut off by the plane at $$x,y$$ and $$z -$$axes respectively.

Hence, for the given equation,

$$a = \frac{5}{2},\;b = 5$$ and $$c = - 5$$

Thus, the intercepts cut off by plane are $$\frac{5}{2},5$$ and $$- 5$$.

## Chapter 11 Ex.11.3 Question 8

Find the equation of the plane with intercept on the $$y$$-axis and parallel to $$ZOX$$ plane.

### Solution

The equation of the plane $$ZOX$$ is $$Y = 0$$

Any plane parallel to it is of the form, $$y = a$$

Since the $$y$$-intercept of the plane is $$3$$,

Therefore, $$a = 3$$

Hence, the equation of the required plane is $$y = 3$$.

## Chapter 11 Ex.11.3 Question 9

Find the equation of the plane through the intersection of the planes $$3x - y + 2z - 4 = 0$$ and $$x + y + z - 2 = 0$$ and the point $$\left( {2,2,1} \right)$$.

### Solution

The equation of the given plane through the intersection of the planes $$3x - y + 2z - 4 = 0$$ and $$x + y + z - 2 = 0$$ is given by

$\left( {3x - y + 2z - 4} \right) + \alpha \left( {x + y + z - 2} \right) = 0;\;\;\alpha \in R \qquad \ldots \left( 1 \right)$

This plane passes through the point $$\left( {2,2,1} \right)$$.

Hence, this point will satisfy equation

\begin{align}& \Rightarrow \left( {3 \times 2 - 2 + 2 \times 1 - 4} \right) + \alpha \left( {2 + 2 + 1 - 2} \right) = 0\\ &\Rightarrow 2 + 3\alpha = 0\\ &\Rightarrow \alpha = \frac{{ - 2}}{3}\end{align}

Putting $$\alpha = \frac{{ - 2}}{3}$$ in equation (1), we get

\begin{align} &\Rightarrow \left( {3x - y + 2z - 4} \right) - \frac{2}{3}\left( {x + y + z - 2} \right) = 0\\& \Rightarrow 3\left( {3x - y + 2z - 4} \right) - 2\left( {x + y + z - 2} \right) = 0\\& \Rightarrow 9x - 3y + 6z - 12 - 2x - 2y - 2z + 4 = 0\\ &\Rightarrow 7x - 5y + 4z - 8 = 0\end{align}

## Chapter 11 Ex.11.3 Question 10

Find the vector equation of the plane passing through the intersection of the planes $$\vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) = 7,\;\vec r.\left( {2\hat i + 5\hat j + 3\hat k} \right) = 9$$ and through the point $$\left( {2,1,3} \right)$$.

### Solution

The equations of the planes are $$\vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) = 7$$ and $$\vec r.\left( {2\hat i + 5\hat j + 3\hat k} \right) = 9$$

Hence,

\begin{align} \Rightarrow \vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) - 7 = 0\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\ \Rightarrow \vec r.\left( {2\hat i + 5\hat j + 3\hat k} \right) - 9 = 0\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Equation of the required plane is given by

$$\left[ {\vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) - 7} \right]{\rm{ }} + \lambda \left[ {\vec r.\left( {2\hat i + 5\hat j + 3\hat k} \right) - 9} \right] = 0;\;\;\lambda \in R$$

Therefore,

\begin{align} &\Rightarrow \vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) - 7{\rm{ }} + \lambda \vec r.\left( {2\hat i + 5\hat j + 3\hat k} \right) - 9\lambda = 0\\ &\Rightarrow \vec r.\left[ {\left( {2\hat i + 2\hat j - 3\hat k} \right) + \lambda \left( {2\hat i + 5\hat j + 3\hat k} \right)} \right] = 9\lambda + 7\\ &\Rightarrow \vec r.\left[ {\left( {2 + 2\lambda } \right)\hat i + \left( {2 + 5\lambda } \right)\hat j + \left( {3\lambda - 3} \right)\hat k]} \right] = 9\lambda + 7\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

The plane passes through the point $$\left( {2,1,3} \right)$$

Hence, its position vector is given by, $$\vec r = 2\hat i + \hat j + 3\hat k$$

Putting in equation (3), we get

\begin{align} &\Rightarrow \left( {2\hat i + \hat j + 3\hat k} \right).\left[ {\left( {2 + 2\lambda } \right)\hat i + \left( {2 + 5\lambda } \right)\hat j + \left( {3\lambda - 3} \right)\hat k]} \right] = 9\lambda + 7\\& \Rightarrow 2\left( {2 + 2\lambda } \right) + \left( {2 + 5\lambda } \right) + 3\left( {3\lambda - 3} \right) = 9\lambda + 7\\ &\Rightarrow 4 + 4\lambda + 2 + 5\lambda + 9\lambda - 9 - 9\lambda - 7 = 0\\& \Rightarrow 9\lambda - 10 = 0\\ &\Rightarrow \lambda = \frac{{10}}{9}\end{align}

Putting $$\lambda = \frac{{10}}{9}$$ in equation (3), we get

\begin{align} &\Rightarrow \vec r.\left( {\frac{{38}}{9}\hat i + \frac{{68}}{9}\hat j + \frac{3}{9}\hat k} \right) = 17\\& \Rightarrow \vec r.\left( {38\hat i + 68\hat j + 3\hat k} \right) = 153\end{align}

## Chapter 11 Ex.11.3 Question 11

Find the equation of the plane through the line of intersection of the planes $$x + y + z = 1$$ and $$2x + 3y + 4z = 5$$ which is perpendicular to the plane $$x - y + z = 0$$.

### Solution

The equation of the plane through the intersection of the planes $$x + y + z = 1$$ and $$2x + 3y + 4z = 5$$is

\begin{align}& \Rightarrow \left( {x + y + z - 1} \right) + \lambda \left( {2x + 3y + 4z - 5} \right) = 0\\ &\Rightarrow \left( {2\lambda + 1} \right)x + \left( {3\lambda + 1} \right)y + \left( {4\lambda + 1} \right)z - \left( {5\lambda + 1} \right) = 0 \qquad \ldots \left( 1 \right)\end{align}

The plane in equation (1) is perpendicular to the plane $$x - y + z = 0$$

Since the planes are perpendicular, $${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0$$

Here,

$${a_1} = \left( {2\lambda + 1} \right),\;{b_1} = \left( {3\lambda + 1} \right)$$ and $${c_1} = \left( {4\lambda + 1} \right)$$

$${a_2} = 1,\;{b_2} = - 1$$ and $${c_2} = 1$$

Hence,

\begin{align} &\Rightarrow \left( {2\lambda + 1} \right) \times 1 + \left( {3\lambda + 1} \right) \times \left( { - 1} \right) + \left( {4\lambda + 1} \right) \times 1 = 0\\& \Rightarrow 2\lambda + 1 - 3\lambda - 1 + 4\lambda + 1 = 0\\& \Rightarrow 3\lambda + 1 = 0\\ &\Rightarrow \lambda = \frac{{ - 1}}{3}\end{align}

Putting $$\lambda = \frac{{ - 1}}{3}$$ in equation (1), we get

\begin{align}&\Rightarrow \frac{1}{3}x + \frac{1}{3}z + \frac{2}{3} = 0\\& \Rightarrow x - z + 2 = 0\end{align}

## Chapter 11 Ex.11.3 Question 12

Find the angle between the planes whose vector equations are $$\vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) = 5$$ and $$\vec r.\left( {3\hat i - 3\hat j + 5\hat k} \right) = 3$$.

### Solution

The equations of the given planes are $$\vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) = 5$$ and $$\vec r.\left( {3\hat i - 3\hat j + 5\hat k} \right) = 3$$

If $${\vec n_1}$$ and $${\vec n_2}$$ are normal to the planes, $$\vec r.{\vec n_1} = {d_1}$$ and $$\vec r.{\vec n_2} = {d_2}$$

Then the angle between them $${\rm{\theta }}$$ is given by,

${\rm{\theta }} = {\cos ^{ - 1}}\frac{{\left| {{{\vec n}_1}.{{\vec n}_2}} \right|}}{{\left| {{{\vec n}_1}} \right|\left| {{{\vec n}_2}} \right|}} \qquad \ldots \left( 1 \right)$

Here,

$${\vec n_1} = 2\hat i + 2\hat j - 3\hat k$$ and $${\vec n_2} = 3\hat i - 3\hat j + 5\hat k$$

Hence,

\begin{align}{{\vec n}_1}.{{\vec n}_2} &= \left( {2\hat i + 2\hat j - 3\hat k} \right).\left( {3\hat i - 3\hat j + 5\hat k} \right)\\ &= 2 \times 3 + 2 \times \left( { - 3} \right) + \left( { - 3} \right) \times 5\\ &= - 15\\\left| {{{\vec n}_1}} \right| &= \sqrt {{{\left( 2 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 3} \right)}^2}} = \sqrt {17} \\\left| {{{\vec n}_2}} \right| &= \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 3} \right)}^2} + {{\left( 5 \right)}^2}} = \sqrt {43}\end{align}

Substituting these values in equation (1), we obtain

\begin{align}{\rm{\theta }} &= {\cos ^{ - 1}}\frac{{\left| { - 15} \right|}}{{\left| {\sqrt {17} } \right|.\left| {\sqrt {43} } \right|}}\\& = {\cos ^{ - 1}}\frac{{15}}{{\sqrt {731} }}\end{align}

## Chapter 11 Ex.11.3 Question 13

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

(a) $$7x + 5y + 6z + 30 = 0$$ and $$3x - y - 10z + 4 = 0$$

(b) $$2x + y + 3z - 2 = 0$$ and $$x - 2y + 5 = 0$$

(c) $$2x - 2y + 4z + 5 = 0$$ and $$3x - 3y + 6z - 1 = 0$$

(d) $$2x - y + 3z - 1 = 0$$ and $$2x - y + 3z + 3 = 0$$

(e) $$4x + 8y + z - 8 = 0$$ and $$y + z - 4 = 0$$

### Solution

The directions ratios of normal to be the plane $${L_1}:{a_1}x + {b_1}y + {c_1}z = 0$$ are $${a_1},{b_1},{c_1}$$ and $${L_2}:{a_2}x + {b_2}y + {c_2}z = 0$$ are $${a_2},{b_2},{c_2}$$

If,

\begin{align}{L_1}\parallel {L_2};\;\; &\Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\\{L_1} \bot {L_2};\;\;& \Rightarrow {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0\end{align}

The angle between $${L_1}$$ and $${L_2}$$ is given by

${\rm{\theta }} = {\cos ^{ - 1}}\left| {\frac{{\left| {{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}} \right|}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} .\sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|$

(a) The equations of the planes are $$7x + 5y + 6z + 30 = 0$$ and $$3x - y - 10z + 4 = 0$$

Here,

$${a_1} = 7,{b_1} = 5$$ and $${c_1} = 6$$

$${a_2} = 3,{b_2} = - 1$$ and $${c_2} = - 10$$

Hence,

\begin{align}{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} &= 7 \times 3 + 5 \times \left( { - 1} \right) + 6 \times \left( { - 10} \right)\\&= - 44\\& \ne 0\end{align}

Therefore, the given planes are not perpendicular.

Also,

\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{7}{3}\\\frac{{{b_1}}}{{{b_2}}} = \frac{5}{{ - 1}} = - 5\\\frac{{{c_1}}}{{{c_2}}} = \frac{6}{{ - 10}} = - \frac{3}{5}\end{align}

It can be seen that, $$\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}$$

Therefore, the given planes are not parallel.

The angle between them is given by,

\begin{align}{\rm{\theta }} &= {\cos ^{ - 1}}\left| {\frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} .\sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|\\ &= {\cos ^{ - 1}}\left| {\frac{{7 \times 3 + 5 \times \left( { - 1} \right) + 6 \times \left( { - 10} \right)}}{{\sqrt {{{\left( 7 \right)}^2} + {{\left( 5 \right)}^2} + {{\left( 6 \right)}^2}} .\sqrt {{{\left( 3 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 10} \right)}^2}} }}} \right|\\&= {\cos ^{ - 1}}\left| {\frac{{21 - 5 - 60}}{{\sqrt {110} .\sqrt {110} }}} \right|\\& = {\cos ^{ - 1}}\left| {\frac{{ - 44}}{{110}}} \right|\\ &= {\cos ^{ - 1}}\frac{2}{5}\end{align}

(b) The equations of the planes are $$2x + y + 3z - 2 = 0$$ and $$x - 2y + 5 = 0$$

Here,

$${a_1} = 2,{b_1} = 1$$ and $${c_1} = 3$$

$${a_2} = 1,{b_2} = - 2$$ and $${c_2} = 0$$

Hence,

\begin{align}{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} &= 2 \times 1 + 1 \times \left( { - 2} \right) + 3 \times 0\\ &= 2 - 2 + 0\\ &= 0\end{align}

Thus, the given planes are perpendicular to each other.

(c) The equations of the planes are $$2x - 2y + 4z + 5 = 0$$ and $$3x - 3y + 6z - 1 = 0$$

Here,

$${a_1} = 2,{b_1} = - 2$$ and $${c_1} = 4$$

$${a_2} = 3,{b_2} = - 3$$ and $${c_2} = 6$$

Hence,

\begin{align}{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} &= 2 \times 3 + \left( { - 2} \right) \times \left( { - 3} \right) + 4 \times 6\\& = 6 + 6 + 24\\ &= 36\\& \ne 0\end{align}

Thus, the given planes are not perpendicular to each other.

Also,

\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{2}{3}\\\frac{{{b_1}}}{{{b_2}}} = \frac{{ - 2}}{{ - 3}} = \frac{2}{3}\\\frac{{{c_1}}}{{{c_2}}} = \frac{4}{6} = \frac{2}{3}\end{align}

It can be seen that, $$\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}$$

Hence, the given planes are parallel to each other.

(d) The equations of the planes are $$2x - y + 3z - 1 = 0$$ and $$2x - y + 3z + 3 = 0$$

Here,

$${a_1} = 2,{b_1} = - 1$$ and $${c_1} = 3$$

$${a_2} = 2,{b_2} = - 1$$ and $${c_2} = 3$$

Hence,

\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{2}{2} &= 1\\\frac{{{b_1}}}{{{b_2}}} &= \frac{{ - 1}}{{ - 1}} = 1\\\frac{{{c_1}}}{{{c_2}}} &= \frac{3}{3} &= 1\end{align}

Therefore, $$\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}$$

Hence, the given lines are parallel to each other.

(e) The equations of the given planes are $$4x + 8y + z - 8 = 0$$ and $$y + z - 4 = 0$$

Here,

$${a_1} = 4,{b_1} = 8$$ and $${c_1} = 1$$

$${a_2} = 0,{b_2} = 1$$ and $${c_2} = 1$$

Hence,

\begin{align}{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} &= 4 \times 0 + 8 \times 1 + 1\\& = 9\\& \ne 0\end{align}

Thus, the given lines are not perpendicular to each other.

Also,

\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{4}{0}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{8}{1} = 8\\\frac{{{c_1}}}{{{c_2}}}& = \frac{1}{1} = 1\end{align}

It can be seen that, $$\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}$$

Thus, the given lines not parallel to each other.

The angle between the planes is given by,

\begin{align}{\rm{\theta }} &= {\cos ^{ - 1}}\left| {\frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} .\sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|\\ &= {\cos ^{ - 1}}\left| {\frac{{4 \times 0 + 8 \times 1 + 1 \times 1}}{{\sqrt {{{\left( 4 \right)}^2} + {{\left( 8 \right)}^2} + {{\left( 1 \right)}^2}} \times \sqrt {{{\left( 0 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} }}} \right|\\ &= {\cos ^{ - 1}}\left| {\frac{9}{{9\sqrt 2 }}} \right|\\ &= {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right)\\& = 45^\circ\end{align}

## Chapter 11 Ex.11.3 Question 14

In the following cases, find the distance of each of the given points from the corresponding given plane.

 Point Plane (a) $$\left( {0,0,0} \right)$$ $$3x - 4y + 12z = 3$$ (b) $$\left( {3, - 2,1} \right)$$ $$2x - y + 2z + 3 = 0$$ (c) $$\left( {2,3, - 5} \right)$$ $$x + 2y - 2z = 9$$ (d) $$\left( { - 6,0,0} \right)$$ $$2x - 3y + 6z - 2 = 0$$

### Solution

The distance between a point, $$P\left( {{x_1},{y_1},{z_1}} \right)$$ and a plane $$Ax + By + Cz + D = 0$$ is given by,

$d = \left| {\frac{{A{x_1} + B{y_1} + C{z_1} + D}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}} \right|$

(a) The given point is $$\left( {0,0,0} \right)$$ and the plane is $$3x - 4y + 12z - 3 = 0$$

Therefore,

\begin{align}d &= \left| {\frac{{3 \times 0 + \left( { - 4} \right) \times 0 + 12 \times 0 + \left( { - 3} \right)}}{{\sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( {12} \right)}^2}} }}} \right|\\ &= \left| {\frac{{ - 3}}{{\sqrt {169} }}} \right|\\ &= \frac{3}{{13}}\end{align}

(b) The given point is $$\left( {3, - 2,1} \right)$$ and the plane is $$2x - y + 2z + 3 = 0$$

Therefore,

\begin{align}d& = \left| {\frac{{2 \times 3 + \left( { - 1} \right) \times \left( { - 2} \right) + 2 \times 1 + 3}}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 2 \right)}^2}} }}} \right|\\ &= \left| {\frac{{13}}{{\sqrt 9 }}} \right|\\ &= \frac{{13}}{3}\end{align}

(c) The given point is $$\left( {2,3, - 5} \right)$$ and the plane is $$x + 2y - 2z - 9 = 0$$

Therefore,

\begin{align}d &= \left| {\frac{{1 \times 2 + 2 \times 3 + \left( { - 2} \right) \times \left( { - 5} \right) + \left( { - 9} \right)}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 2} \right)}^2}} }}} \right|\\ &= \left| {\frac{9}{{\sqrt 9 }}} \right|\\ &= \frac{9}{3}\\&= 3\end{align}

(d) The given point is $$\left( { - 6,0,0} \right)$$ and the plane is $$2x - 3y + 6z - 2 = 0$$

Therefore,

\begin{align}d &= \left| {\frac{{2 \times \left( { - 6} \right) + \left( { - 3} \right) \times 0 + 6 \times 0 + \left( { - 2} \right)}}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( { - 3} \right)}^2} + {{\left( 6 \right)}^2}} }}} \right|\\ &= \left| {\frac{{ - 14}}{{\sqrt {49} }}} \right|\\ &= \frac{{14}}{7}\\ &= 2\end{align}

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