# NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry (Ex 11.3) Exercise 11.3

## NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry (Ex 11.3) Exercise 11.3

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## NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.3

Exercise 11.3

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

a. $z = 2$

Ans: The equation of the plane is $z = 2$ or $0x + 0y + z = 2\,\,\,\,\,\,\,\,.....(1)$

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

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

Now, dividing both sides of equation (1) by 1, we get

$\Rightarrow \left( {\frac{0}{1}} \right).x + \left( {\frac{0}{1}} \right).y + \left( {\frac{1}{1}} \right).z = 2$

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

$lx + my + nz = d$is the form of the equation.

Therefore, $l = 0$, $m = 0$, $n = 1$ and $d = 2$.

$l,m,n$ are direction cosine normal to the plane.

d, the distance of the perpendicular drawn from the origin.

Here, $0,0$and $1$are the direction cosines.

The distance is $2$units.

b. $x + y + z = 1$

Ans: Let $x + y + z = 1\,\,\,\,\,\,.....(1)$

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

$\therefore \sqrt {{1^2} + {1^2} + {1^2}} = \sqrt 3$

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

$\frac{1}{{\sqrt 3 }}x + \frac{1}{{\sqrt 3 }}y + \frac{1}{{\sqrt 3 }}z = \frac{1}{{\sqrt 3 }}\,\,\,\,.....(2)$

$lx + my + nz = d$is the form of the above equation.

$l,m,n$ are direction cosine normal to the plane.

d, the distance of normal.

Here$\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}$ are direction cosines.

The distance is $\frac{1}{{\sqrt 3 }}$units.

c. $2x + 3y - z = 5$

Ans: Let, $2x + 3y - z = 5\,\,\,\,\,.....(1)$

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

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

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

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

$lx + my + nz = d$is the form of the above equation.

$l,m,n$are directional cosine normal to the plane.

$d$, the distance of normal.

Here$\frac{2}{{\sqrt {14} }},\frac{3}{{\sqrt {14} }}$, and $\frac{{ - 1}}{{\sqrt {14} }}$, are the direction cosines.

The distance is $\frac{5}{{\sqrt {14} }}$units.

d. $5y + 8 = 0$

Ans: The equation is $5y + 8 = 0$

$\Rightarrow 0x - 5y + 0z = 8\,\,\,\,\,\,.....(1)$

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

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

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

$- y = \frac{8}{5}$

$lx + my + nx = d$is the form of the above equation.

$l,m,n$are directional cosine normal to the plane.

$d$, the distance from the origin.

Here $0, - 1,0$are direction cosines.

The distance is $\frac{8}{5}$units.

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

Ans: The normal vector is, $\mathop n\limits^ \to = 3\mathop i\limits^ \wedge + 5\mathop j\limits^ \wedge - 6\mathop k\limits^ \wedge$

$\mathop n\limits^ \wedge = \frac{{\mathop n\limits^ \to }}{{\left| {\mathop n\limits^ \to } \right|}} = \frac{{3\mathop i\limits^ \wedge + 5\mathop j\limits^ \wedge - 6\mathop k\limits^ \wedge }}{{\sqrt {{3^2} + {5^2} + {{\left( { - 6} \right)}^2}} }} = \frac{{3\mathop i\limits^ \wedge + 5\mathop j\limits^ \wedge - 6\mathop k\limits^ \wedge }}{{\sqrt {70} }}$

Knowing that the equation of the plane with position vector $\mathop r\limits^ \to$is given by, $\mathop r\limits^ \to .\mathop n\limits^ \wedge = d$

$\Rightarrow \mathop r\limits^ \wedge .\left( {\frac{{3\mathop i\limits^ \wedge + 5\mathop j\limits^ \wedge - 6\mathop k\limits^ \wedge }}{{\sqrt {70} }}} \right) = 7$

Hence this is the vector equation of the required plane.

3. Find the Cartesian equation of the following planes:

a). $\mathop r\limits^ \to .\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge - \mathop k\limits^ \wedge } \right) = 2$

Ans: Given that the equation of the plane is

$\mathop r\limits^ \to .\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge - \mathop k\limits^ \wedge } \right) = 2\,\,\,\,\,.....(1)$

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

$\mathop r\limits^ \to = x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge$

Substituting $\mathop r\limits^ \to$in  (1), we obtain

$\left( {x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge } \right).\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge - \mathop k\limits^ \wedge } \right) = 2$

$\Rightarrow x + y - z = 2$

Hence this is the Cartesian equation of the plane.

b). $\mathop r\limits^ \to .\left( {2\mathop i\limits^ \wedge + 3\mathop j\limits^ \wedge - 4\mathop k\limits^ \wedge } \right) = 1$

Ans:

Let, $\mathop r\limits^ \to .\left( {2\mathop i\limits^ \wedge + 3\mathop j\limits^ \wedge - 4\mathop k\limits^ \wedge } \right) = 1\,\,\,\,\,\,.....(1)$

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

$\mathop r\limits^ \to = x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge$

Substituting $\mathop r\limits^ \to$in equation (1), we obtain

$\left( {x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge } \right).\left( {2\mathop i\limits^ \wedge + 3\mathop j\limits^ \wedge - 4\mathop k\limits^ \wedge } \right) = 1$

$\Rightarrow 2x + 3y - 4z = 1$

Hence this is the Cartesian equation of the plane.

c). $\mathop r\limits^ \to .\left( {\left( {s - 2t} \right)\mathop i\limits^ \wedge + \left( {3 - t} \right)\mathop j\limits^ \wedge + \left( {2s + t} \right)\mathop k\limits^ \wedge } \right) = 15\,\,$

Ans:

Let, $\mathop r\limits^ \to .\left( {\left( {s - 2t} \right)\mathop i\limits^ \wedge + \left( {3 - t} \right)\mathop j\limits^ \wedge + \left( {2s + t} \right)\mathop k\limits^ \wedge } \right) = 15\,\,\,\,\,.....(1)$

For any arbitrary point $P\left( {x,y,z} \right)$on the plane, position vectors $\mathop r\limits^ \to$are given by,

$\mathop r\limits^ \to = x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge$

Substituting $\mathop r\limits^ \to$in equation (1), we obtain

$\left( {x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge } \right).\left( {\left( {s - 2t} \right)\mathop i\limits^ \wedge + \left( {3 - t} \right)\mathop j\limits^ \wedge + \left( {2s + t} \right)\mathop k\limits^ \wedge } \right) = 15$

$\Rightarrow \left( {s - 2t} \right)x + \left( {3 - t} \right)y + \left( {2s + t} \right)z = 15$

Hence this is the Cartesian equation of the given plane.

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

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

Ans:

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

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

$\Rightarrow 2x + 3y + 4z = 12\,\,\,\,\,\,.....(1)$

The direction ratios of normal are $2,3$and $4$

$\sqrt {{2^2} + {3^2} + {4^2}} = \sqrt {29}$

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

$\frac{2}{{\sqrt {29} }}x + \frac{3}{{\sqrt {29} }}y + \frac{4}{{\sqrt {29} }}z = \frac{{12}}{{\sqrt {29} }}$

$lx + my + nz = d$is the form of the equation

$l,m,n$are direction cosines

Distance of normal is $d$ from the origin.

The foot of the perpendicular is $\left( {ld,md,nd} \right)$.

So, here the foot of the perpendicular is

$\left( {\frac{2}{{\sqrt {29} }} \times \frac{{12}}{{\sqrt {29} }},\frac{3}{{\sqrt {29} }} \times \frac{{12}}{{\sqrt {29} }},\frac{4}{{\sqrt {29} }} \times \frac{{12}}{{\sqrt {29} }}} \right) = \left( {\frac{{24}}{{29}},\frac{{36}}{{29}},\frac{{48}}{{29}}} \right)$

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

Ans:

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

$3x + 4y - 6 = 0$

$\Rightarrow 0x + 3y + 4z = 6\,\,\,\,\,.....(1)$

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

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

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

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

$lx + my + nz = d$is the form of the equation

$l,m,n$are direction cosines

Distance of normal is $d$ from the origin.

The foot of the perpendicular is $\left( {ld,md,nd} \right)$.

So here, the foot of the perpendicular is,

$\left( {0,\frac{3}{5} \times \frac{6}{5},\frac{4}{5} \times \frac{6}{5}} \right) = \left( {0,\frac{{18}}{{25}},\frac{{24}}{{25}}} \right)$

(iii). $x + y + z = 1$

Ans:

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

$x + y + z = 1\,\,\,\,\,......(1)$

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

$\therefore \sqrt {{1^2} + {1^2} + {1^2}} = \sqrt 3$

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

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

$lx + my + nz = d$ is the form of the equation

$l,m,n$are direction cosines

Distance of normal is $d$ from the origin.

The foot of the perpendicular is $\left( {ld,md,nd} \right)$.

So here, the foot of the perpendicular is,

$\left( {\frac{1}{{\sqrt 3 }} \times \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }} \times \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }} \times \frac{1}{{\sqrt 3 }}} \right) = \left( {\frac{1}{3},\frac{1}{3},\frac{1}{3}} \right)$

(iv). $5y + 8 = 0$

Ans:

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

$5y + 8 = 0$

$\Rightarrow 0x - 5y + 0z = 8\,\,\,\,\,\,\,\,......(1)$

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

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

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

$- y = \frac{8}{5}$

$lx + my + nz = d$ is the form of the equation

$l,m,n$are direction cosines.

Distance of normal is $d$ from the origin.

The foot of the perpendicular is $\left( {ld,md,nd} \right)$.

So here, the foot of the perpendicular is,

$\left( {0, - 1 \times \left( {\frac{8}{5}} \right),0} \right) = \left( {0, - \left( {\frac{8}{5}} \right),0} \right)$

5. Find the vector and cartesian equations of the planes

A. that passes through the point  $\left( {1,0, - 2} \right)$ and the normal to the plane is$\mathop i\limits^ \wedge + \mathop j\limits^ \wedge - \mathop k\limits^ \wedge$

Ans: The position vector of the point $\left( {1,0, - 2} \right)$is $\mathop a\limits^ \to = \mathop i\limits^ \wedge - 2\mathop k\limits^ \wedge$

$\mathop N\limits^ \to = \mathop i\limits^ \wedge + \mathop j\limits^ \wedge - \mathop k\limits^ \wedge$ is the normal vector which is perpendicular to the plane.

Vector equation of the plane is, $\left( {\mathop r\limits^ \to - \mathop a\limits^ \to } \right).\mathop N\limits^ \to = 0$

$\Rightarrow \left[ {\mathop r\limits^ \to - \left( {\mathop i\limits^ \wedge - \mathop {2k}\limits^ \wedge } \right)} \right].\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge - \mathop k\limits^ \wedge } \right) = 0\,\,\,\,\,\,\,......(1)$

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

$\therefore \mathop r\limits^ \to$ = $x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge$

Therefore, equation (1) becomes

$\left[ {\left( {x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge } \right) - \left( {\mathop i\limits^ \wedge - \mathop {2k}\limits^ \wedge } \right)} \right].\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge - \mathop k\limits^ \wedge } \right) = 0\,$

$\Rightarrow \left( {\left( {x - 1} \right)\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + \left( {z + 2} \right)\mathop k\limits^ \wedge } \right).\left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge - \mathop k\limits^ \wedge } \right) = 0\,$

$\Rightarrow \left( {x - 1} \right) + y - \left( {z + 2} \right) = 0$

$\Rightarrow x + y - z - 3 = 0$

$\Rightarrow x + y - z = 3$

Hence this is the Cartesian equation of the required plane.

B. that passes through the point $\left( {1,4,6} \right)$ and the normal vector to the plane is $\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + \mathop k\limits^ \wedge$

Ans: The position vector of the point $\left( {1,4,6} \right)$ is $\mathop a\limits^ \to = \mathop i\limits^ \wedge + 4\mathop j\limits^ \wedge + 6\mathop k\limits^ \wedge$

$\mathop N\limits^ \to = \mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + \mathop k\limits^ \wedge$ is the normal vector which is perpendicular to the plane.

Vector equation of the plane is, $\left( {\mathop r\limits^ \to - \mathop a\limits^ \to } \right).\mathop N\limits^ \to = 0$

$\Rightarrow \left[ {\mathop r\limits^ \to - \left( {\mathop i\limits^ \wedge + 4\mathop j\limits^ \wedge + 6\mathop k\limits^ \wedge } \right)} \right].\left( {\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + \mathop k\limits^ \wedge } \right) = 0\,\,\,\,\,\,\,......(1)$

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

$\therefore \mathop r\limits^ \to = x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge$

Therefore, equation (1) becomes

$\left[ {\left( {x\mathop i\limits^ \wedge + y\mathop j\limits^ \wedge + z\mathop k\limits^ \wedge } \right) - \left( {\mathop i\limits^ \wedge + 4\mathop j\limits^ \wedge + 6\mathop k\limits^ \wedge } \right)} \right].\left( {\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + \mathop k\limits^ \wedge } \right) = 0\,$

$\Rightarrow \left( {\left( {x - 1} \right)\mathop i\limits^ \wedge + \left( {y - 4} \right)\mathop j\limits^ \wedge + \left( {z - 6} \right)\mathop k\limits^ \wedge } \right).\left( {\mathop i\limits^ \wedge - 2\mathop j\limits^ \wedge + \mathop k\limits^ \wedge } \right) = 0\,$

$\Rightarrow \left( {x - 1} \right) - 2\left( {y - 4} \right) + \left( {z - 6} \right) = 0$

$\Rightarrow x - 2y + z + 1 = 0$

Hence this is the Cartesian equation of the required plane.

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

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

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

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

$\Rightarrow 2 + 2 - 4 = 0$

The points $A,B,C$are collinear, and so there will be an infinite number of planes passing through the given points.

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

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

$\left| {\begin{array}{*{20}{c}} 1&1&0 \\ 1&2&1 \\ { - 2}&2&{ - 1} \end{array}} \right| = \left( { - 2 - 2} \right) - \left( {-1 + 2} \right) = - 5$

$\Rightarrow - 5\ne 0$

A plane will therefore pass through the point $A,B,C$

It is known that the equation of the plane through the points,$\left( {{x_1},{y_1},{z_1}} \right),\left( {{x_2},{y_2},{z_2}} \right),\left( {{x_3},{y_3},{z_3}} \right)$is,

$\left| {\begin{array}{*{20}{c}} {x - {x_1}}&{y - {y_1}}&{z - {z_1}} \\ {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}} \\ {{x_3} - {x_1}}&{{y_3} - {y_1}}&{{z_3} - {z_1}} \end{array}} \right| = 0$

$\Rightarrow \left| {\begin{array}{*{20}{c}} {x - 1}&{y - 1}&z \\ 0&1&1 \\ { - 3}&1&{ - 1} \end{array}} \right| = 0$

$\Rightarrow - 2\left( {x - 1} \right) - 3\left( {y - 1} \right) + 3z = 0$

$\Rightarrow - 2x - 3y + 3z + 2 + 3 = 0$

$\Rightarrow - 2x - 3y + 3z = - 5$

$\Rightarrow 2x + 3y - 3z = 5$

Hence this is the Cartesian equation of the required plane.

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

Ans: $2x + y - z = 5\,\,\,\,.....(1)$

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

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

$\Rightarrow \frac{x}{{\frac{5}{2}}} + \frac{y}{5} + \frac{z}{{ - 5}} = 1\,\,\,\,\,......(2)$

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, is the plane in intercept form

The intercepts cut off are$a,b,c$ respectively at$x,y,z$

Therefore, for the given equation,

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

The intercepts cut off are $\frac{5}{2},5, - 5$

8. Find the equation of the plane with intercept $3$ on the y-axis and parallel to $ZOX$ plane.

Ans: The equation of the plane $ZOX$is $y = 0$

The plane parallel to it is, $y = a$

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

$\therefore a = 3$

Thus, the equation of the plane is $y = 3$

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 $\left( {2,2,1} \right)$.

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

$\left( {3x - y + 2z - 4} \right) + \alpha \left( {x + y + z - 2} \right) = 0,$where $\alpha \in R\,\,\,\,....\left( 1 \right)$

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

Therefore, this point will safety equation (1)

$\therefore \left( {\left( {3 \times 2} \right) - 2 + \left( {2 \times 1} \right) - 4} \right) + \alpha \left( {2 + 2 + 1 - 2} \right) = 0$

$\Rightarrow 2 + 3\alpha = 0$

$\Rightarrow \alpha = - \frac{2}{3}$

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

$\left( {3x - y + 2z - 4} \right) - \frac{2}{3}\left( {x + y + z - 2} \right) = 0$

$\Rightarrow 3\left( {3x - y + 2z - 4} \right) - 2\left( {x + y + z - 2} \right) = 0$   $\Rightarrow \left( {9x - 3y + 6z - 12} \right) - 2\left( {x + y + z - 2} \right) = 0$

$\Rightarrow 7x - 5y + 4z - 8 = 0$

This is the equation of the plane.

10. Find the vector equation of the plane passing through the intersection of the planes $\mathop r\limits^ \to .\left( {2\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge - 3\mathop k\limits^ \wedge } \right) = 7,\mathop r\limits^ \to .\left( {2\mathop i\limits^ \wedge + 5\mathop j\limits^ \wedge + 3\mathop k\limits^ \wedge } \right) = 9$ and through the point $\left( {2,1,3} \right)$.

Ans: The equations of the planes are $\mathop r\limits^ \to .\left( {2\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge - 3\mathop k\limits^ \wedge } \right) = 7$ and $\mathop r\limits^ \to .\left( {2\mathop i\limits^ \wedge + 5\mathop j\limits^ \wedge + 3\mathop k\limits^ \wedge } \right) = 9$

$\Rightarrow \mathop r\limits^ \to .\left( {2\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge - 3\mathop k\limits^ \wedge } \right) - 7 = 0\,\,\,\,\,......(1)$

$\Rightarrow \mathop r\limits^ \to .\left( {2\mathop i\limits^ \wedge + 5\mathop j\limits^ \wedge + 3\mathop k\limits^ \wedge } \right) - 9 = 0\,\,\,\,\,......(2)$

So the equation of the plane is,

$\Rightarrow \left[ {\mathop r\limits^ \to .\left( {2\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge - 3\mathop k\limits^ \wedge } \right) - 7} \right] + \lambda \left[ {\mathop r\limits^ \to .\left( {2\mathop i\limits^ \wedge + 5\mathop j\limits^ \wedge + 3\mathop k\limits^ \wedge } \right) - 9} \right] = 0\,\,$ where $\lambda \in R$

$\Rightarrow \mathop r\limits^ \to .\left[ {\left( {2\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge - 3\mathop k\limits^ \wedge } \right) + \lambda \left( {2\mathop i\limits^ \wedge + 5\mathop j\limits^ \wedge + 3\mathop k\limits^ \wedge } \right)} \right] = 9\lambda + 7$

$\Rightarrow \mathop r\limits^ \to .\left( {\left( {2 + 2\lambda } \right)\mathop i\limits^ \wedge + \left( {2 + 5\lambda } \right)\mathop j\limits^ \wedge + \left( {3\lambda - 3} \right)\mathop k\limits^ \wedge } \right) = 9\lambda + 7\,\,\,\,\,\,\,\,\,\,\,......(3)$

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

The position vector is given by,

$\mathop r\limits^ \to = 2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + 3\mathop k\limits^ \wedge$

Substituting in equation (3), we obtain

$\Rightarrow \left( {2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + 3\mathop k\limits^ \wedge } \right).\left( {\left( {2 + 2\lambda } \right)\mathop i\limits^ \wedge + \left( {2 + 5\lambda } \right)\mathop j\limits^ \wedge + \left( {3\lambda - 3} \right)\mathop k\limits^ \wedge } \right) = 9\lambda + 7$

$\Rightarrow 2\left( {2 + 2\lambda } \right) + 1\left( {2 + 5\lambda } \right) + 3\left( {3\lambda - 1} \right) = 9\lambda + 7$

$\Rightarrow 4 + 4\lambda + 2 + 5\lambda + 9\lambda - 9 = 9\lambda + 7$

$\Rightarrow \lambda = \frac{{10}}{9}$

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

$\Rightarrow \mathop r\limits^ \to .\left( {\frac{{38}}{9}\mathop i\limits^ \wedge + \frac{{68}}{9}\mathop j\limits^ \wedge + \frac{3}{9}\mathop k\limits^ \wedge } \right) = 17$

$\Rightarrow \mathop {\mathop r\limits^ \to .\left( {38\mathop i\limits^ \wedge + 68\mathop j\limits^ \wedge + 3\mathop k\limits^ \wedge } \right) = 153}\limits^ \to$

Hence this is the equation of the required plane.

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 $x - y + z = 0$.

Ans: The equation of the plane through

$x + y + z = 1$and $2x + 3y + 4z = 5$is,

$\left( {x + y + z - 1} \right) + \lambda \left( {2x + 3y + 4z - 5} \right) = 0$

$\Rightarrow \left( {2\lambda + 1} \right)x + \left( {3\lambda + 1} \right)y + \left( {4\lambda + 1} \right)z - \left( {5\lambda + 1} \right) = 0\,\,\,\,.....(1)$

The direction ratios, ${a_{1,}}{b_1},{c_1}$of this plane, are $\left( {2\lambda + 1} \right),\left( {3\lambda + 1} \right)$, and $\left( {4\lambda + 1} \right)$

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

Its direction ratios ${a_{2,}}{b_2},{c_2}$are $1, - 1,1$. Since the planes are perpendicular,

${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0$

$\Rightarrow \left( {2\lambda + 1} \right) - \left( {3\lambda + 1} \right) + \left( {4\lambda + 1} \right) = 0$

$\Rightarrow 3\lambda + 1 = 0$

$\Rightarrow \lambda = - \frac{1}{3}$

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

$\frac{1}{3}x - \frac{1}{3}z + \frac{2}{3} = 0$

$\Rightarrow x - z + 2 = 0$

12. Find the angle between the planes whose vector equations are $\mathop {\mathop r\limits^ \to .\left( {2\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge - 3\mathop k\limits^ \wedge } \right)}\limits^ \to = 5$and $\mathop {\mathop r\limits^ \to .\left( {3\mathop i\limits^ \wedge - 3\mathop j\limits^ \wedge + 5\mathop k\limits^ \wedge } \right)}\limits^ \to = 3$

Ans: The equation of the given planes are $\mathop {\mathop r\limits^ \to .\left( {2\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge - 3\mathop k\limits^ \wedge } \right)}\limits^ \to = 5$ and $\mathop {\mathop r\limits^ \to .\left( {3\mathop i\limits^ \wedge - 3\mathop j\limits^ \wedge + 5\mathop k\limits^ \wedge } \right)}\limits^ \to = 3$

It is known that if $\mathop {{n_1}}\limits^ \to$and ${\mathop n\limits^ \to _2}$is normal to the planes, $\mathop r\limits^ \to .\mathop {{{\mathop n\limits^ \to }_1} = {d_1}}\limits^{}$and $\mathop r\limits^ \to .\mathop {{n_2}}\limits^ \to = {d_2}$,

Normal to the planes, then the angle between them, $\theta$, is given by,

$\cos \theta = \left| {\frac{{{{\mathop n\limits^ \to }_1}.\mathop {{n_2}}\limits^ \to }}{{\left| {{{\mathop n\limits^ \to }_1}} \right|\left| {\mathop {{n_2}}\limits^ \to } \right|}}} \right|\,\,\,\,\,\,\,\,\,\,\,......(1)$

Here, $\mathop {{n_1}}\limits^ \to = 2\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge - 3\mathop k\limits^ \wedge$and $\mathop {{n_2}}\limits^ \to = 3\mathop i\limits^ \wedge - 3\mathop j\limits^ \wedge + 5\mathop k\limits^ \wedge$

$\therefore \mathop {{n_1}.}\limits^ \to \mathop {{n_2}}\limits^ \to = \left( {2\mathop i\limits^ \wedge + 2\mathop j\limits^ \wedge - 3\mathop k\limits^ \wedge } \right)\left( {3\mathop i\limits^ \wedge - 3\mathop j\limits^ \wedge + 5\mathop k\limits^ \wedge } \right) = \left( {2 \times 3} \right) + \left( {2 \times - 3} \right) + \left( { - 3 \times 5} \right) = - 15$

$\left| {\mathop {{n_1}}\limits^ \to } \right| = \sqrt {{2^2} + {2^2} + {{\left( { - 3} \right)}^2}} = \sqrt {17}$

$\left| {\mathop {{n_2}}\limits^ \to } \right| = \sqrt {{3^2} + {{\left( { - 3} \right)}^2} + {5^2}} = \sqrt {43}$

Substituting in equation (1) we obtain

$\Rightarrow \cos \theta = \left| {\frac{{ - 15}}{{\sqrt {17} .\sqrt {43} }}} \right|$

$\Rightarrow \cos \theta = \frac{{15}}{{\sqrt {731} }}$

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

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.

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

Ans: 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 ${a_2},{b_2},{c_2}$

The planes are parallel i.e. ${L_1}\parallel {L_2}$if $\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}$

The planes are perpendicular, i.e. ${L_1} \bot {L_2}$if ${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0$

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

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

The equation of the plane is, $7x + 5y + 6z + 30 = 0$ and $3x - y - 10z + 4 = 0$

Here,

$\begin{gathered} {a_1} = 7,{b_1} = 5,{c_1} = 6 \hfill \\ {a_2} = 3,{b_2} = - 1,{c_2} = - 10 \hfill \\ \end{gathered}$

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

So, planes are not perpendicular.

$\frac{{{a_1}}}{{{a_2}}} = \frac{7}{3},\frac{{{b_1}}}{{{b_2}}} = \frac{5}{{ - 1}} = - 5,\frac{{{c_1}}}{{{c_2}}} = \frac{6}{{ - 10}} = \frac{{ - 3}}{5}$

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

So, the planes are not parallel.

The angle between them is given by,

$\theta = {\cos ^{ - 1}}\left| {\frac{{7 \times 3 + 5 \times \left( { - 1} \right) + 6 \times \left( { - 10} \right)}}{{\sqrt {{7^2} + {5^2} + {6^2}\sqrt {{3^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 10} \right)}^2}} } }}} \right|$

$\Rightarrow {\cos ^{ - 1}}\left| {\frac{{21 - 5 - 60}}{{\sqrt {110 \times \sqrt {110} } }}} \right| = {\cos ^{ - 1}}\frac{{44}}{{110}}$

$\theta = {\cos ^{ - 1}}\frac{2}{5}$

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

Ans: 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 ${a_2},{b_2},{c_2}$

The planes are parallel i.e. ${L_1}\parallel {L_2}$if $\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}$

The planes are perpendicular, i.e. ${L_1} \bot {L_2}$if ${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0$

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

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

The equation of the plane is, $2x + y + 3z - 2 = 0$and $x - 2y + 5 = 0$

Here,

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

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

$\Rightarrow {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = \left( {2 \times 1} \right) + \left( {1 \times - 2} \right) + \left( {3 \times 0} \right) = 0$

So, the planes are perpendicular.

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

Ans: 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 ${a_2},{b_2},{c_2}$

The planes are parallel i.e. ${L_1}\parallel {L_2}$if $\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}$

The planes are perpendicular, i.e. ${L_1} \bot {L_2}$if ${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0$

The equation of the plane is $2x - 2y + 4z + 5 = 0$and $3x - 3y + 6z - 1 = 0$

Here,

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

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

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

So, given planes are not perpendicular.

$\frac{{{a_1}}}{{{a_2}}} = \frac{2}{3},\frac{{{b_1}}}{{{b_2}}} = \frac{{ - 2}}{{ - 3}} = \frac{2}{3},\frac{{{c_1}}}{{{c_2}}} = \frac{4}{6} = \frac{2}{3}$

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

So, the planes are parallel.

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

Ans:

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 ${a_2},{b_2},{c_2}$

The planes are parallel i.e. ${L_1}\parallel {L_2}$if $\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}$

The planes are perpendicular, i.e. ${L_1} \bot {L_2}$if ${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0$

The equation of the planes are $2x - y + 3z - 1 = 0$and $2x - y + 3z + 3 = 0$

Here,

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

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

$\frac{{{a_1}}}{{{a_2}}} = \frac{2}{2} = 1,\frac{{{b_1}}}{{{b_2}}} = \frac{{ - 1}}{{ - 1}} = 1,\frac{{{c_1}}}{{{c_2}}} = \frac{3}{3} = 1$

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

So, the planes are parallel.

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

Ans:

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 ${a_2},{b_2},{c_2}$

The planes are parallel i.e. ${L_1}\parallel {L_2}$if $\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}$

The planes are perpendicular, i.e. ${L_1} \bot {L_2}$if ${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0$

The equation of the  is plane, $4x + 8y + z - 8 = 0$and $y + z - 4 = 0$

Here,

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

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

$\Rightarrow {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = \left( {4 \times 0} \right) + \left( {8 \times 1} \right) + \left( {1 \times 1} \right) = 9 \ne 0$

So, they are not perpendicular.

$\frac{{{a_1}}}{{{a_2}}} = \frac{4}{0},\frac{{{b_1}}}{{{b_2}}} = \frac{8}{1} = 8,\frac{{{c_1}}}{{{c_2}}} = \frac{1}{1} = 1$

$\therefore \frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}$

So, they are not parallel.

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

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

$\theta = {\cos ^{ - 1}}\left| {\frac{{4 \times 0 + 8 \times \left( 1 \right) + 1 \times \left( 1 \right)}}{{\sqrt {{4^2} + {8^2} + {1^2} \times \sqrt {{0^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} } }}} \right| = {\cos ^{ - 1}}\left| {\frac{9}{{9\sqrt 2 }}} \right| = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right)$

$\therefore \theta = {45^0}$

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) $\left( { - 6,0,0} \right)$ $2x - 3y + 6z - 2 = 0$

a). Point $\left( {0,0,0} \right)$and plane $3x - 4y + 12z = 3$

Ans:

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

$d = \left| {\frac{{A{x_1} + B{y_1} + C{z_1} - D}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}} \right|\,\,\,\,\,\,......(1)$

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

$\therefore d = \left| {\frac{{3 \times 0 - 4 \times 0 + 12 \times 0 - 3}}{{\sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( {12} \right)}^2}} }}} \right|\, = \frac{3}{{\sqrt {169} }} = \frac{3}{{13}}$

b). Point $\left( {3, - 2,1} \right)$and plane $2x - y + 2z + 3 = 0$

Ans:

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

$d = \left| {\frac{{A{x_1} + B{y_1} + C{z_1} - D}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}} \right|\,\,\,\,\,\,......(1)$

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

$\therefore d = \left| {\frac{{\left( {2 \times 3} \right) + \left( { - 2 \times - 1} \right) + \left( {1 \times 2} \right) + 3}}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 2 \right)}^2}} }}} \right|\, = \left| {\frac{{13}}{3}} \right| = \frac{{13}}{3}$

c). Point $\left( {2,3, - 5} \right)$and plane $x + 2y - 2z = 9$

Ans:

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

$d = \left| {\frac{{A{x_1} + B{y_1} + C{z_1} - D}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}} \right|\,\,\,\,\,\,......(1)$

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

$\therefore d = \left| {\frac{{\left( {1 \times 2} \right) + \left( {3 \times 2} \right) + \left( { - 2 \times - 5} \right) - 9}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 2} \right)}^2}} }}} \right|\, = \frac{9}{3} = 3$

d). Point $\left( { - 6,0,0} \right)$and plane $2x - 3y + 6z - 2 = 0$

Ans:

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

$d = \left| {\frac{{A{x_1} + B{y_1} + C{z_1} - D}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}} \right|\,\,\,\,\,\,......(1)$

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

$\therefore d = \left| {\frac{{\left( {2 \times - 6} \right) + \left( { - 3 \times 0} \right) + \left( {6 \times 0} \right) - 2}}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( { - 3} \right)}^2} + {{\left( 6 \right)}^2}} }}} \right|\, = \left| {\frac{{ - 14}}{{\sqrt {49} }}} \right| = \frac{{14}}{7} = 2$

## NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.3

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