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

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

PDF of NCERT Solutions for Class 12 Math Chapter 11 Exercise 11.3 prepared by expert Mathematics teacher at Vedantu as per CBSE (NCERT) books guidelines. Download our Class 12 Math Chapter 11 Three Dimensional Geometry Ex 11.3 Questions with Solutions pdf by clicking once on the pdf link given below to help you to revise the complete syllabus and score More marks in your exams.

 Class: NCERT Solutions for Class 12 Subject: Class 12 Maths Chapter Name: Chapter 11 - Three-Dimensional Geometry Exercise: Exercise - 11.3 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

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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

Opting for the NCERT solutions for Ex 11.3 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.3 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 11 Exercise 11.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 12 Maths Chapter 11 Exercise 11.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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## Some of The Topics Covered In Class 12 Chapter 11 Exercise 11.3

1. Plane

1. Equation of a plane in normal form

2. Equation of a plane perpendicular to a given vector, passing through given point

3. Equation of a plane passing through three non collinear points

4. Intercept form of the equation of a plane

5. Plane passing through the intersection of two given planes

2. Coplanarity of Two Lines

3. The angle between Two Planes

4. Distance of a Point from a Plane

5. The angle between a Line and a Plane

### NCERT Solutions for Class 11 Maths Chapters

Chapter 11 - Three-Dimensional Geometry

## FAQs on NCERT Solutions for Class 12 Maths Chapter 11: Three Dimensional Geometry - Exercise 11.3

1. What is a plane?

A two-dimensional surface is another name for a plane. A plane has an infinite length, an infinite width, and no curvature. The flat surfaces of a cube or cuboid, as well as the flat surface of the paper, are all examples of real-world geometric planes; however, it can be challenging to imagine a plane in everyday life.

2. What is the coplanarity of two lines?

When two lines in a three-dimensional space are located on the same plane, they are said to be coplanar. We now know how to visualise the equation of a line in three dimensions using vector notation.

3. How Do You Refer To the Angle Between Two Planes?

A dihedral angle is the angle formed when two planes cross. To put it another way, the dihedral angle is the angle formed by the normal vectors of any two planes.

4. What do you mean by Three-dimensional geometry?

The three coordinates in the XYZ plane—x, y, and z—are used in 3-Dimensional geometry, which deals with the mathematics of shapes in 3-D space. 3D shapes are those that take up physical space. Solid shapes with three dimensions—height, width, and length—are also known as 3D shapes. All known matter exists in three-dimensional space, a geometric three-parameter model with three axes (x, y, and z). The length, width, height, depth, and breadth of the phrase are the three dimensions that were selected.

5. Why should one opt for the NCERT solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry (Ex 11.3) Exercise 11.3 by Vedantu?

When it comes to exam preparation, the NCERT solutions for Class 12 Maths Ex 11.3, from Vedantu are thought to be the best choice for CBSE students. There are numerous exercises in this chapter. On this page, in PDF format, we have the Exercise 11.3 Class 12 Maths NCERT solutions. This solution is available for download at your convenience, or you can access it directly from the Vedantu website or app to study it.