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 (Ex 11.3) Exercise 11.3

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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 Exercise 11.3

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

The chapter on Three Dimensional Geometry is one of the most important chapters from an examination point of view. Various essential concepts are covered in Ex 11.3 Class12 of this chapter. Students learn to find the equation of a plane. These equations are mostly found in vector form. However, at times they are also found in the Cartesian form. Significant concepts like the equation of a plane in normal form, the equation of a plane perpendicular to a given vector and passing through a given point, the equation of a plane passing through three non-collinear points, a plane passing through the intersection of two planes, the angle between two planes, the distance between a point and a plane are discussed in this chapter. Students should try to solve all the 14 questions in Exercise 11.3 Class 12 Maths. For problems, they are facing difficulty to solve, they should refer to the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3 provided by Extramarks. The questions have been solved easily and simply in the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3 so that students can easily solve such questions in the examination.

Q.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) 2x + 3y – z = 5 (d) 5y + 8 = 0

Ans

$\begin{array}{l} (a) The equatin of the plane is z=2 \Rightarrow 0 .x+0 .y+z=2 … (i) \\ The direction ratios of normal are 0, 0 and 1 . \\ ∴ \sqrt{0^{2} {+0}^{2} {+1}^{2}} =1 \\ So, according to rule, dividing equation â€„ (i) by 1, we get \\ 0 .x+0 .y+z=2, which is in the form of lx+my+nz=d, \\ where direction cosines, l=0,m=0 and n=1 and the \\ distance of the plane from origin is 2 units . \\ (b) x+y+z=1 … (i) \\ The direction ratios of normal are 1, 1 and 1 . \\ \sqrt{1^{2} {+1}^{2} {+1}^{2}} = \sqrt{3} \\ Dividing equation (i) by \sqrt{3}, we get \\ \frac{x}{\sqrt{3}} + \frac{y}{\sqrt{3}} + \frac{z}{\sqrt{3}} = \frac{1}{\sqrt{3}} … (ii) \\ Comparing equation (ii) with the equation lx+my+nz=d,we get \\ Direction cosines: \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} and the distance of plane from \\ the origin is \frac{1}{\sqrt{3}} units . \\ (c) 2x+3y-z=5 … (i) \\ The direction ratios of normal are 2, 3 and-1 . \\ \sqrt{2^{2} {+3}^{2} + {(-1)}^{2}} = \sqrt{14} \\ Dividing equation (i) by \sqrt{14}, we get \\ \frac{2x}{\sqrt{14}} + \frac{3y}{\sqrt{14}} – \frac{z}{\sqrt{14}} = \frac{5}{\sqrt{14}} … (ii) \\ Comparing equation (ii) with the equation lx+my+nz=d,we get \\ Direction cosines: \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}} and the distance of plane from \\ the origin is \frac{5}{\sqrt{14}} units . \\ (d) 5y+8=0 \\ \Rightarrow -0x-5y-0 z=8 … (i) \\ The direction ratios of normal are 0, -5 and 0 . \\ \sqrt{0^{2} + {(-5)}^{2} {+0}^{2}} =5 \\ Dividing equation (i) by 5, we get \\ – \frac{0x}{5} – \frac{5y}{5} – \frac{0z}{5} = \frac{8}{5} … (ii) \\ Comparing equation (ii) with the equation lx+my+nz=d,we get \\ Direction cosines: 0,-1,0 and the distance of plane from \\ the origin is \frac{8}{5} units . \end{array}$

Q.2 Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector

$3 \hat{i} + 5 \hat{j} - 6 \hat{k} .$

Ans

$\begin{array}{l} Here, \vec{n} = 3 \hat{i} + 5 \hat{j} - 6 \hat{k} \\ \hat{n} = \frac{\vec{n}}{|\vec{n}|} \\ = \frac{3 \hat{i} + 5 \hat{j} - 6 \hat{k}}{|3 \hat{i} + 5 \hat{j} - 6 \hat{k}|} \\ = \frac{3 \hat{i} + 5 \hat{j} - 6 \hat{k}}{\sqrt{3^{2} + 5^{2} + {(- 6)}^{2}}} \\ = \frac{3 \hat{i} + 5 \hat{j} - 6 \hat{k}}{\sqrt{9 + 25 + 36}} \\ = \frac{3 \hat{i} + 5 \hat{j} - 6 \hat{k}}{\sqrt{70}} \\ The equation of the plane with position vector \vec{r} is given by \\ \vec{r} . \hat{n} = d \\ \vec{r} (\frac{3 \hat{i} + 5 \hat{j} - 6 \hat{k}}{\sqrt{70}}) = 7 \\ This is the vector eqution of â€‹ the plane . \end{array}$

Q.3

$\begin{array}{l} Find the cartesian equation of the following â€‹ planes : \\ (a) \vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2 \\ (b) \vec{r} . (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1 \\ (c) \vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15 \end{array}$

Ans

$\begin{array}{l} (a) The given equation of the plane is \\ \vec{r} (\hat{i} + \hat{j} - \hat{k}) = 2 . .. (i) \\ For any arbitrary point P (x, y, z) on the plane, position vector \\ \vec{r} is given by, \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} . \\ Substitutin value of \vec{r} in equation (i), we get \\ (x \hat{i} + y \hat{j} + z \hat{k}) (\hat{i} + \hat{j} - \hat{k}) = 2 \\ x + y - z = 2 \\ This is the cartesian equation of the plane . \\ (b) The given equation of the plane is \\ \vec{r} (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1 . .. (i) \\ For any arbitrary point P (x, y, z) on the plane, position vector \\ \vec{r} is given by, \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \\ Substitutin value of \vec{r} in equation (i), we get \\ (x \hat{i} + y \hat{j} + z \hat{k}) (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 1 \\ 2 x + 3 y - 4 z = 1 \\ This is the cartesian equation of the plane . \\ (c) The given equation of the plane is \\ \vec{r} . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 15 . .. (i) \\ For any arbitrary point P (x, y, z) on the plane, position vector \\ \vec{r} is given by, \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \\ Substitutin value of \vec{r} in equation (i), we get \\ (x \hat{i} + y \hat{j} + z \hat{k}) . [(s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}] = 1 \\ (s - 2 t) x + (3 - t) y + (2 s + t) z = 1 \\ This is the cartesian equation of the plane . \end{array}$

Q.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

Ans

$\begin{array}{l} (a) T h e g i v e n e q u a t i o n o f t h e p l a n e i s \\ 2 x +3 y +4 z -12=0 \\ \Rightarrow 2 x +3 y +4 z =12 . .. (i) \\ L e t t h e c o o r d i n a t e s o f t h e f o o t o f p e r p e n d i c u l a r f r o m o r i g i n t o \\ t h e p l a n e i s (x_{1}, y_{1}, z_{1}) . T h e n p o i n t (x_{1}, y_{1}, z_{1}) w i l l s a t i s f y \\ e q u a t i o n (i), s o \\ 2 x_{1} + 3 y_{1} + 4 z_{1} = 12 \\ T h e n, t h e d i r e c t i o n r a t i o s o f t h e p e r p e n d i c u l a r a r e 2, 3, 4 . \\ D i v i d i n g e q u a t i o n (i) b y \sqrt{2^{2} + 3^{2} + 4^{2}} = \sqrt{29}, w e g e t \\ \Rightarrow \frac{2}{\sqrt{29}} x + \frac{3}{\sqrt{29}} y + \frac{4}{\sqrt{29}} z = \frac{12}{\sqrt{29}} \\ T h i s e q u a t i o n i s i n t h e f o r m o f l x + m y + n z = d, w h e r e l, m a n d n \\ a r e t h e d i r e c t i o n c o s i n e s o f n o r m a l t o t h e p l a n e a n d d i s t h e \\ d i s t a n c e o f t h e n o r m a l f r o m t h e o r i g i n . \\ T h e c o o r d i n a t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e g i v e n b y \\ (l d, m d, n d) . \\ T h e r e f o r e, t h e c o o r d i a n t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e \\ (\frac{2}{\sqrt{29}} . \frac{12}{\sqrt{29}}, \frac{3}{\sqrt{29}} . \frac{12}{\sqrt{29}}, \frac{4}{\sqrt{29}} . \frac{12}{\sqrt{29}}) = (\frac{24}{29}, \frac{36}{29}, \frac{48}{29}) . \\ (b) T h e g i v e n e q u a t i o n o f t h e p l a n e i s \\ 3 y +4 z -6=0 \\ \Rightarrow 0 . x +3 y +4 z =12 . .. (i) \\ L e t t h e c o o r d i n a t e s o f t h e f o o t o f p e r p e n d i c u l a r f r o m o r i g i n t o \\ t h e p l a n e i s (x_{1}, y_{1}, z_{1}) . T h e n p o i n t (x_{1}, y_{1}, z_{1}) w i l l s a t i s f y \\ e q u a t i o n (i), s o \\ 0 . x_{1} + 3 y_{1} + 4 z_{1} = 6 \\ T h e n, t h e d i r e c t i o n r a t i o s o f t h e p e r p e n d i c u l a r a r e 0, 3, 4 . \\ D i v i d i n g e q u a t i o n (i) b y \sqrt{0^{2} + 3^{2} + 4^{2}} = 5, w e g e t \\ \Rightarrow \frac{0}{5} x + \frac{3}{5} y + \frac{4}{5} z = \frac{6}{5} \\ T h i s e q u a t i o n i s i n t h e f o r m o f l x + m y + n z = d, w h e r e l, m a n d n \\ a r e t h e d i r e c t i o n c o s i n e s o f n o r m a l t o t h e p l a n e a n d d i s t h e \\ d i s t a n c e o f t h e n o r m a l f r o m t h e o r i g i n . \\ T h e c o o r d i n a t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e g i v e n b y \\ (l d, m d, n d) . \\ T h e r e f o r e, t h e c o o r d i a n t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e \\ (\frac{0}{5} . \frac{6}{5}, \frac{3}{5} . \frac{6}{5}, \frac{4}{5} . \frac{6}{5}) = (0, \frac{18}{25}, \frac{24}{25}) . \\ (c) T h e g i v e n e q u a t i o n o f t h e p l a n e i s \\ x + y + z =1 . .. (i) \\ L e t t h e c o o r d i n a t e s o f t h e f o o t o f p e r p e n d i c u l a r f r o m o r i g i n t o \\ t h e p l a n e i s (x_{1}, y_{1}, z_{1}) . T h e n p o i n t (x_{1}, y_{1}, z_{1}) w i l l s a t i s f y \\ e q u a t i o n (i), s o \\ x + y + z =1 \\ T h e n, t h e d i r e c t i o n r a t i o s o f t h e p e r p e n d i c u l a r a r e 1, 1, 1 . \\ D i v i d i n g e q u a t i o n (i) b y \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3}, w e g e t \\ \Rightarrow \frac{1}{\sqrt{3}} x + \frac{1}{\sqrt{3}} y + \frac{1}{\sqrt{3}} z = \frac{1}{\sqrt{3}} \\ T h i s e q u a t i o n i s i n t h e f o r m o f l x + m y + n z = d, w h e r e l, m a n d n \\ a r e t h e d i r e c t i o n c o s i n e s o f n o r m a l t o t h e p l a n e a n d d i s t h e \\ d i s t a n c e o f t h e n o r m a l f r o m t h e o r i g i n . \\ T h e c o o r d i n a t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e g i v e n b y \\ (l d, m d, n d) . \\ T h e r e f o r e, t h e c o o r d i a n t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e \\ (\frac{1}{\sqrt{3}} . \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} . \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} . \frac{1}{\sqrt{3}}) = (\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) . \\ (d) T h e g i v e n e q u a t i o n o f t h e p l a n e i s \\ 5 y +8=0 \\ 0 . x -5 y +0 . z = 8 . .. (i) \\ L e t t h e c o o r d i n a t e s o f t h e f o o t o f p e r p e n d i c u l a r f r o m o r i g i n t o \\ t h e p l a n e i s (x_{1}, y_{1}, z_{1}) . T h e n p o i n t (x_{1}, y_{1}, z_{1}) w i l l s a t i s f y \\ e q u a t i o n (i), s o \\ 0 . x_{1} - 5 y_{1} + 0 . z_{1} = 8 \\ T h e n, t h e d i r e c t i o n r a t i o s o f t h e p e r p e n d i c u l a r a r e 0, 5, 0 . \\ D i v i d i n g e q u a t i o n (i) b y \sqrt{0^{2} + {(- 5)}^{2} + 0^{2}} = 5, w e g e t \\ \Rightarrow \frac{0}{5} x - \frac{5}{5} y + \frac{0}{5} z = \frac{8}{5} \\ T h i s e q u a t i o n i s i n t h e f o r m o f l x + m y + n z = d, w h e r e l, m a n d n \\ a r e t h e d i r e c t i o n c o s i n e s o f n o r m a l t o t h e p l a n e a n d d i s t h e \\ d i s t a n c e o f t h e n o r m a l f r o m t h e o r i g i n . \\ T h e c o o r d i n a t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e g i v e n b y \\ (l d, m d, n d) . \\ T h e r e f o r e, t h e c o o r d i a n t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e \\ (0 \times \frac{8}{5}, \frac{- 5}{5} \times \frac{8}{5}, 0 \times \frac{8}{5}) = (0,- \frac{8}{5}, 0) . \end{array}$

Q.5

$\begin{array}{l} Find the vector and cartesian equations of the planes \\ (a) that passes through the point (1, 0, - 2) and the normal \\ to the plane is \hat{i} + \hat{j} - \hat{k} . \\ (b) that passes through the point (1, 4, 6) and the normal \\ vector to the plane is \hat{i} - 2 \hat{j} + \hat{k} . \end{array}$

Ans

$\begin{array}{l} (a) The position vector of point (1, 0, - 2), \vec{a} = \hat{i} + 0 . \hat{j} - 2 \hat{k} and the \\ normal vector \vec{N} perpendicular to the plane as \vec{N} = \hat{i} + \hat{j} - \hat{k} \\ Therefore, the vector equation ofthe plane is given by \\ (\vec{r} - \vec{a}) . \vec{N} = 0 \\ or [\vec{r} - (\hat{i} + 0 . \hat{j} - 2 \hat{k})] . (\hat{i} + \hat{j} - \hat{k}) = 0 . .. (i) \\ \vec{r} is the position vector of any point P (x, y, z) in the plane . \\ ∴ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \\ Therefore, equation (i) becomes \\ [x \hat{i} + y \hat{j} + z \hat{k} - (\hat{i} + 0 . \hat{j} - 2 \hat{k})] . (\hat{i} + \hat{j} - \hat{k}) = 0 \\ \Rightarrow [(x - 1) \hat{i} + y \hat{j} + (z + 2) \hat{k}] . (\hat{i} + \hat{j} - \hat{k}) = 0 \\ \Rightarrow (x - 1) + y - (z + 2) = 0 \\ \Rightarrow x + y - z = 3 \\ This is the cartesian equation of the â€‹ required plane . \\ (b) The position vector of point â€‹ (1, 4, 6), \vec{a} = \hat{i} + 4 \hat{j} + 6 \hat{k} and the \\ normal vector \vec{N} perpendicular to the plane as \vec{N} = \hat{i} - 2 \hat{j} + \hat{k} \\ Therefore, the vector equation of the plane is given by \\ (\vec{r} - \vec{a}) . \vec{N} = 0 \\ or [\vec{r} - (\hat{i} + 4 \hat{j} + 6 \hat{k})] . (\hat{i} - 2 \hat{j} + \hat{k}) = 0 . .. (i) \\ \vec{r} is the position vector of any point P (x, y, z) in the plane . \\ ∴ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \\ Therefore, equation (i) becomes \\ [x \hat{i} + y \hat{j} + z \hat{k} - (\hat{i} + 4 \hat{j} + 6 \hat{k})] . (\hat{i} - 2 \hat{j} + \hat{k}) = 0 \\ \Rightarrow [(x - 1) \hat{i} + (y - 4) \hat{j} + (z - 6) \hat{k}] . (\hat{i} - 2 \hat{j} + \hat{k}) = 0 \\ \Rightarrow (x - 1) - 2 (y - 4) + (z - 6) = 0 \\ \Rightarrow x - 2 y + z + 1 = 0 \\ This is the cartesian equation of the required plane . \end{array}$

Q.6 Find the equations of the planes that passes through three points.
(a) (1, 1, – 1), (6, 4, –5), (–4, –2, 3)

(b) (1, 1, 0), (1, 2, 1), (–2, 2, –1)

Ans

$\begin{array}{l} (a) L e t vector forms of the given three points are: \\ \vec{a} = \hat{i} + \hat{j} - \hat{k}, \vec{b} = 6 \hat{i} + 4 \hat{j} - 5 \hat{k} and \vec{c} = - 4 \hat{i} - 2 \hat{j} + 3 \hat{k} \\ T h e n, the vector equation of the plane passing through \vec{a}, \vec{b} and \\ \vec{c} is given by (\vec{r} - \vec{a}) . [(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})] = 0 \\ i . e ., [\vec{r} - (\hat{i} + \hat{j} - \hat{k})] . [(5 \hat{i} + 3 \hat{j} - 4 \hat{k}) \times (- 5 \hat{i} - 3 \hat{j} + 4 \hat{k})] = 0 \\ \Rightarrow [\vec{r} - (\hat{i} + \hat{j} - \hat{k})] . | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 3 & - 4 \\ - 5 & - 3 & 4 \end{matrix} | = 0 \\ \Rightarrow [\vec{r} - (\hat{i} + \hat{j} - \hat{k})] . (0. \hat{i} - 0 \hat{. j} + 0. \hat{k}) = 0 \\ \Rightarrow 0 = 0 \\ S i n c e, the given three points are collinear, so there will be \\ infininte numbers of planes passing through the given points . \\ (b) L e t vector forms of the given three points are: \\ \vec{a} = \hat{i} + \hat{j} + 0 \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} and \vec{c} = - 2 \hat{i} + 2 \hat{j} - \hat{k} \\ T h e n, the vector equation of the plane passing through \vec{a}, \vec{b} and \\ \vec{c} is given by (\vec{r} - \vec{a}) . [(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})] = 0 \\ i . e ., [\vec{r} - (\hat{i} + \hat{j})] . [(0 \hat{i} + \hat{j} + \hat{k}) \times (- 3 \hat{i} + \hat{j} - \hat{k})] = 0 \\ \Rightarrow [\vec{r} - (\hat{i} + \hat{j})] . | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ - 3 & 1 & - 1 \end{matrix} | = 0 \\ \Rightarrow [x \hat{i} + y \hat{j} + z \hat{k} - (\hat{i} + \hat{j})] . (- 2 \hat{i} - 3 \hat{j} + 3 \hat{k}) = 0 \\ \Rightarrow [(x - 1) \hat{i} + (y - 1) \hat{j} + z \hat{k}] . (- 2 \hat{i} - 3 \hat{j} + 3 \hat{k}) = 0 \\ \Rightarrow - 2 (x - 1) - 3 (y - 1) + 3 z = 0 \\ \Rightarrow - 2 x + 2 - 3 y + 3 + 3 z = 0 \\ \Rightarrow 2 x + 3 y - 3 z = 5 \\ T h i s is the cartesian equation of the required plane . \end{array}$

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

Ans

$\begin{array}{l} The given equation of the plane is : \\ 2 x + y - z = 5 . .. (i) \\ Dividing both sides of equation (i) by 5, we get \\ \frac{2 x}{5} + \frac{y}{5} - \frac{z}{5} = \frac{5}{5} \\ \Rightarrow \frac{x}{(\frac{5}{2})} + \frac{y}{5} + \frac{z}{(- 5)} = 1 \\ Comparing this equation of the plane with \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1, we get \\ intercepts cut by plane at x, y and z axes : \\ a = \frac{5}{2}, b = 5 and c = - 5 \\ Thus, the intercepts cut off by the plane are \frac{5}{2}, 5 and - 5 . \end{array}$

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

Ans

The equation of ZOX plane is y=0

Equation of any plane parallel to ZOX plane is

y = a

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

a = 3

Thus, the equation of the required plane is y = 3.

Q.9 Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 with (2, 2, 1).

Ans

$\begin{array}{l} T h e e q u a t i o n s o f g i v e n p l a n e s a r e : \\ 3 x - y + 2 z - 4 = 0 a n d x + y + z - 2 = 0 \\ T h e e q u a t i o n o f p l a n e t h r o u g h i n t e r s e c t i o n o f g i v e n p l a n e s i s \\ (3 x - y + 2 z - 4) + λ (x + y + z - 2) = 0 w h e r e λ \in R … (i) \\ T h e p l a n e p a s s e s t h r o u g h t h e p o i n t (2, 2, 1), s o f r o m e q u a t i o n (i) \\ (3 \times 2 - 2 + 2 \times 1 - 4) + λ (2 + 2 + 1 - 2) = 0 \\ 2 + λ (3) = 0 \Rightarrow λ = - \frac{2}{3} \\ P u t t i n g v a l u e o f λ i n e q u a t i o n (i), w e g e t \\ (3 x - y + 2 z - 4) - \frac{2}{3} (x + y + z - 2) = 0 \\ \Rightarrow 3 (3 x - y + 2 z - 4) - 2 (x + y + z - 2) = 0 \\ \Rightarrow 9 x - 3 y + 6 z - 12 - 2 x - 2 y - 2 z + 4 = 0 \\ \Rightarrow 7 x - 5 y + 4 z - 8 = 0 \\ This is the required equation of the plane . \end{array}$

Q.10

$\begin{array}{l} Find the vector equation of the plane passing through the \\ inter section of the planes \vec{r} . (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 7, \\ \vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9 â€„ and â€„ through â€„ the â€„ point â€„ (2, 1, 3) . \end{array}$

Ans

$\begin{array}{l} The givenen equations of the plane are : \\ \vec{r} . (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 7 or \vec{r} . (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) - 7 = 0 . .. (i) \\ \vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9 or \vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) - 9 = 0 . .. (ii) \\ The equation of any plane throught the intersection of the \\ planes given in equation (i) and (ii) is given by \\ {\vec{r} . (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) - 7} + λ {\vec{r} . (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) - 9} = 0, λ \in R \\ \vec{r} . [(2 \hat{i} + 2 \hat{j} - 3 \hat{k}) + λ (2 \hat{i} + 5 \hat{j} + 3 \hat{k})] = 9 λ + 7 \\ \vec{r} . [(2 + 2 λ) \hat{i} + (2 + 5 λ) \hat{j} - (3 - 3 λ) \hat{k}] = 9 λ + 7 . .. (iii) \\ The plane passes through the point (2, 1, 3) . Therefore, its \\ position vector is given by, \\ \vec{r} = 2 \hat{i} + \hat{j} + 3 \hat{k} \\ Substituting value of \vec{r} in equation (iii), we get \\ (2 \hat{i} + \hat{j} + 3 \hat{k}) . [(2 + 2 λ) \hat{i} + (2 + 5 λ) \hat{j} - (3 - 3 λ) \hat{k}] = 9 λ + 7 \\ \Rightarrow 2 (2 + 2 λ) + (2 + 5 λ) - 3 (3 - 3 λ) = 9 λ + 7 \\ \Rightarrow 4 + 4 λ + 2 + 5 λ - 9 + 9 λ = 9 λ + 7 \\ \Rightarrow 9 λ - 3 = 7 \\ \Rightarrow λ = \frac{10}{9} \\ Putting λ = \frac{10}{9} in equation (iii), we get \\ \vec{r} . [(2 + 2 \times \frac{10}{9}) \hat{i} + (2 + 5 \times \frac{10}{9}) \hat{j} - (3 - 3 \times \frac{10}{9}) \hat{k}] = 9 \times \frac{10}{9} + 7 \\ \vec{r} . [38 \hat{i} + 68 \hat{j} + 3 \hat{k}] = 153 \\ This is the vector equation of the required plane . \end{array}$

Q.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

$\begin{array}{l} T h e e q u a t i o n s o f g i v e n p l a n e s a r e : \\ x + y + z = 1 a n d 2 x + 3 y + 4 z = 5 \\ T h e e q u a t i o n o f p l a n e t h r o u g h i n t e r s e c t i o n o f g i v e n p l a n e s i s \\ (x + y + z - 1) + λ (2 x + 3 y + 4 z - 5) = 0 w h e r e λ \in R \\ \Rightarrow (1 + 2 λ) x + (1 + 3 λ) y + (1 + 4 λ) z = 5 λ + 1 … (i) \\ T h e direction ratios of the plane are (1 + 2 λ), (1 + 3 λ) and (1 + 4 λ) . \\ P l a n e (i) is perpendicular to x - y + z = 0, whose direction ratios \\ are 1, - 1, 1. \\ S i n c e, plane (i) and (ii) are perpendicular, so \\ a_{1} a_{2} {+b}_{1} b_{2} {+c}_{1} c_{2} = 0 \\ \Rightarrow (1 + 2 λ) (1) + (1 + 3 λ) (- 1) + (1 + 4 λ) (1) = 0 \\ \Rightarrow 1 + 2 λ - 1 - 3 λ + 1 + 4 λ = 0 \\ \Rightarrow 3 λ = - 1 \\ \Rightarrow λ = - \frac{1}{3} \\ S u b s t i t u t i n g â€‹ λ = - \frac{1}{3} in equation (i), we get \\ \Rightarrow (1 + 2 \times - \frac{1}{3}) x + (1 + 3 \times - \frac{1}{3}) y + (1 + 4 \times - \frac{1}{3}) z = 5 \times - \frac{1}{3} + 1 \\ \Rightarrow \frac{1}{3} x + (0) y - \frac{1}{3} z = - \frac{2}{3} \\ x - z = - 2 \\ x - z + 2 = 0 \\ T h i s is the required equation of the plane . \\ T h e direction ratios of the plane are (1 + 2 λ), (1 + 3 λ) and (1 + 4 λ) . \\ P l a n e (i) is perpendicular to x - y + z = 0, whose direction ratios \\ are 1, - 1, 1. \\ S i n c e, plane (i) and (ii) are perpendicular, so \\ a_{1} a_{2} {+b}_{1} b_{2} {+c}_{1} c_{2} = 0 \\ \Rightarrow (1 + 2 λ) (1) + (1 + 3 λ) (- 1) + (1 + 4 λ) (1) = 0 \\ \Rightarrow 1 + 2 λ - 1 - 3 λ + 1 + 4 λ = 0 \\ \Rightarrow 3 λ = - 1 \\ \Rightarrow λ = - \frac{1}{3} \\ S u b s t i t u t i n g â€‹ λ = - \frac{1}{3} in equation (i), we get \\ \Rightarrow (1 + 2 \times - \frac{1}{3}) x + (1 + 3 \times - \frac{1}{3}) y + (1 + 4 \times - \frac{1}{3}) z = 5 \times - \frac{1}{3} + 1 \\ \Rightarrow \frac{1}{3} x + (0) y - \frac{1}{3} z = - \frac{2}{3} \\ x - z = - 2 \\ x - z + 2 = 0 \\ T h i s is the required equation of the plane . \end{array}$

Q.12

$\begin{array}{l} Find â€„ the â€„ angle â€„ between â€„ the â€„ planes â€„ whose â€„ vector â€„ equations \\ are â€„ \vec{r} (2 \hat{i} + 2 \hat{j} - \hat{k}) = 5 â€„ and â€„ \vec{r} (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = \end{array}$

Ans

$\begin{array}{l} Theplaneswhosevectorequationsare \\ \vec{r} (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 5 and \vec{r} (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 3 \\ comparing with \vec{r} . \vec{n_{1}} = d_{1} and \vec{r} . \vec{n_{2}} = d_{2}, then \\ \vec{n_{1}} = (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) and \vec{n_{2}} = (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) \\ ∴ \vec{n_{1}} . \vec{n_{2}} = (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) . (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) \\ = 6 - 6 - 15 \\ = - 15 \\ | \vec{n_{1}} | = | 2 \hat{i} + 2 \hat{j} - 3 \hat{k} | \\ = \sqrt{2^{2} + 2^{2} + {(- 3)}^{2}} \\ = \sqrt{4 + 4 + 9} = \sqrt{17} \\ | \vec{n_{2}} | = | 3 \hat{i} - 3 \hat{j} + 5 \hat{k} | \\ = \sqrt{3^{2} + {(- 3)}^{2} + 5^{2}} \\ = \sqrt{9 + 9 + 25} = \sqrt{43} \\ Let angle between two planes be θ, then \\ cosθ = \frac{\vec{n_{1}} . \vec{n_{2}}}{| \vec{n_{1}} | | \vec{n_{2}} |} \\ = \frac{- 15}{\sqrt{17} \sqrt{43}} \\ = - \frac{15}{\sqrt{731}} \\ θ = \cos^{- 1} (- \frac{15}{\sqrt{731}}) \end{array}$

Q.13 In the following cases, determine whether
the given planes are parallel or perpendicular and in case they are neither, find the angle 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 +6 z – 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

Ans

$\begin{array}{l} (a) 7 x + 5 y + 6 z + 30 = 0 a n d 3 x - y - 10 z + 4 = 0 \\ D i r e c t i o n ratios of the given planes are: \\ a_{1} = 7, b_{1} = 5, c_{1} = 6 \\ a_{2} = 3, b_{2} = - 1, c_{2} = - 10 \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 7 \times 3 + 5 \times - 1 + 6 \times - 10 \\ = 21 - 5 - 60 = - 44 \neq 0 \\ Therefore, the given planes are not perpendicular to each other . \\ \frac{a_{1}}{a_{2}} = \frac{7}{3}, \frac{b_{1}}{b_{2}} = \frac{5}{- 1} and \frac{c_{1}}{c_{2}} = \frac{6}{- 10} = - \frac{3}{5} \\ \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ Therefore, the given planes are not parallel to each other . \\ The angle between two planes is \\ θ = {cos}^{- 1} |(\frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}})| \end{array}$ $\begin{array}{l} = {cos}^{- 1} |(\frac{7 \times 3 + 5 \times - 1 + 6 \times - 10}{\sqrt{7^{2} + 5^{2} + 6^{2}} \sqrt{3^{2} + {(- 1)}^{2} + {(- 10)}^{2}}})| \\ = {cos}^{- 1} |(\frac{- 44}{\sqrt{110} \sqrt{110}})| \\ = {cos}^{- 1} (\frac{44}{110}) \\ = {cos}^{- 1} (\frac{2}{5}) \end{array}$ $\begin{array}{l} (b) 2 x + y + 3 z - 2 = 0 a n d x - 2 y + 5 = 0 \\ D i r e c t i o n ratios of the given planes are: \\ a_{1} = 2, b_{1} = 1, c_{1} = 3 \\ a_{2} = 1, b_{2} = - 2, c_{2} = 0 \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 \times 1 + 1 \times - 2 + 3 \times 0 \\ = 2 - 2 + 0 = 0 \\ Therefore, the given planes are perpendicular to each other . \\ (c) 2 x - 2 y + 4 z + 5 = 0 a n d 3 x - 3 y + 6 z - 1 = 0 \\ D i r e c t i o n ratios of the given planes are: \\ a_{1} = 2, b_{1} = - 2, c_{1} = 4 \\ a_{2} = 3, b_{2} = - 3, c_{2} = 6 \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 \times 3 + - 2 \times - 3 + 4 \times 6 \\ = 6 + 6 + 24 = 36 \neq 0 \end{array}$ $\begin{array}{l} Therefore, the given planes are not perpendicular to each other . \\ \frac{a_{1}}{a_{2}} = \frac{2}{3}, \frac{b_{1}}{b_{2}} = \frac{- 2}{- 3} = \frac{2}{3} and \frac{c_{1}}{c_{2}} = \frac{4}{6} = \frac{2}{3} \end{array}$ $\begin{array}{l} \Rightarrow \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ Therefore, the given planes are parallel to each other . \\ (d) 2 x - y + 3 z - 1 = 0 a n d 2 x - y + 3 z + 3 = 0 \\ D i r e c t i o n ratios of the given planes are: \\ a_{1} = 2, b_{1} = - 1, c_{1} = 3 \\ a_{2} = 2, b_{2} = - 1, c_{2} = 3 \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 \times 2 + - 1 \times - 1 + 3 \times 3 \\ = 4 + 1 + 9 = 14 \neq 0 \\ Therefore, the given planes are not perpendicular to each other . \\ \frac{a_{1}}{a_{2}} = \frac{2}{2} = 1, \frac{b_{1}}{b_{2}} = \frac{- 1}{- 1} = 1 and \frac{c_{1}}{c_{2}} = \frac{3}{3} = 1 \\ \Rightarrow \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ Therefore, the given planes are parallel to each other . \\ (e) 4 x + 8 y + z - 8 = 0 a n d y + z - 4 = 0 \\ D i r e c t i o n ratios of the given planes are: \\ a_{1} = 4, b_{1} = 8, c_{1} = 1 \\ a_{2} = 0, b_{2} = 1, c_{2} = 1 \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 4 \times 0 + 8 \times 1 + 1 \times 1 \\ = 0 + 8 + 1 = 9 \neq 0 \end{array}$ $\begin{array}{l} Therefore, the given planes are not perpendicular to each other . \\ \frac{a_{1}}{a_{2}} = \frac{4}{0}, \frac{b_{1}}{b_{2}} = \frac{8}{1} = 8 and \frac{c_{1}}{c_{2}} = \frac{1}{1} = 1 \\ \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \end{array}$ $\begin{array}{l} Therefore, the given planes are not parallel to each other . \\ The angle between two planes is \\ θ = {cos}^{- 1} |(\frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}})| \\ = {cos}^{- 1} |(\frac{4 \times 0 + 8 \times 1 + 1 \times 1}{\sqrt{4^{2} + 8^{2} + 1^{2}} \sqrt{0^{2} + 1^{2} + 1^{2}}})| \\ = {cos}^{- 1} |(\frac{9}{\sqrt{16 + 64 + 1} \sqrt{0 + 1 + 1}})| \\ = {cos}^{- 1} |(\frac{9}{\sqrt{81} \sqrt{2}})| \\ = {cos}^{- 1} |\frac{1}{\sqrt{2}}| = \frac{π}{4} \\ T h u s, the angle between two planes is \frac{π}{4} . \end{array}$

Q.14 In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(a) (0, 0, 0) 3x – 4y + 12 z = 3

(b) (3, –2, 1) 2x – y + 2z + 3 = 0

(d) (–6, 0, 0) 2x – 3y + 6z – 2 =0

Ans

$\begin{array}{l} (a) The given equation of plane is 3 x - 4 y + 12 z = 3 and point \\ is â€„ (0, 0, 0) . \\ T h e distance of plane from given point (d) = |\frac{3 (0) - 4 (0) + 12 (0) - 3}{\sqrt{3^{2} + {(- 4)}^{2} + {(12)}^{2}}}| \\ = \frac{3}{\sqrt{9 + 16 + 144}} \\ = \frac{3}{\sqrt{169}} \\ = \frac{3}{13} \\ (b) The given equation of plane is 2 x - y + 2 z + 3 = 0 and point \\ is â€„ (3, - 2, 1) . \\ T h e distance of plane from given point (d) = |\frac{2 (3) - (- 2) + 2 (1) + 3}{\sqrt{2^{2} + {(- 1)}^{2} + {(2)}^{2}}}| \\ = \frac{13}{\sqrt{4 + 1 + 4}} \\ = \frac{13}{\sqrt{9}} \\ = \frac{13}{3} \\ (c) â€„ The given equation of plane is x - 2 y + 2 z - 9 = 0 and point \\ is â€„ (2, 3, - 5) . \\ T h e distance of plane from given point (d) = |\frac{(2) - 2 (3) + 2 (- 5) - 9}{\sqrt{1^{2} + {(- 2)}^{2} + {(2)}^{2}}}| \\ = \frac{23}{\sqrt{1 + 4 + 4}} \\ = \frac{23}{\sqrt{9}} \\ = \frac{23}{3} \\ (d) â€„ T h e â€„ g i v e n â€„ e q u a t i o n â€„ o f â€„ p l a n e â€„ i s â€„ 2 x - 3 y + 6 z - 2 = 0 â€„ a n d â€„ p o i n t \\ i s â€„ (- 6, 0, 0) . \\ T h e â€„ d i s t a n c e â€„ o f p l a n e â€„ f r o m â€„ g i v e n â€„ p o i n t â€„ (d) â€„ = |\frac{2 (- 6) - 3 (0) + 6 (0) - 2}{\sqrt{2^{2} + {(- 3)}^{2} + 6^{2}}}| \\ = \frac{14}{\sqrt{4 + 9 + 36}} \\ = \frac{14}{\sqrt{49}} = \frac{14}{7} = 2 \end{array}$

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FAQs (Frequently Asked Questions)

1. What topics are covered in the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3?

The NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3 cover the topic of Planes and their equations.

2. What is the number of questions in the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3?

There are 14 questions in Exercise 11.3. Solutions are given for all the 14 questions in the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3.

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Students looking for the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3 can find it in PDF format on the Extramarks website and mobile application. Practising these questions before the board exams helps boost students’ performance in the board exams. Before the board exams, students should practice as many NCERT questions as they can and refer to the solutions on Extramarks including the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3. This will give them the necessary experience to solve the questions in the exam.

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The NCERT textbook on Mathematics is widely regarded as the most important book for CBSE board exams. Questions are frequently framed directly from NCERT exercises and examples. The NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3 provide answers to these questions clearly. Correctly answering these questions may help students earn higher marks.

5. Do NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3 help students understand the concepts covered in the chapter?

When studying the chapter is complete, the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3 are a good way to practice. The step-by-step solutions in the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3 can assist students with concept clarification. Apart from that, Extramarks’ various study materials and learning modules can assist students in maximising their preparation.

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

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

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

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

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FAQs (Frequently Asked Questions)

1. What topics are covered in the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3?

2. What is the number of questions in the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3?

3. Where can a student find the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3?

4. How do the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3 help students score higher in CBSE board exams?

5. Do NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3 help students understand the concepts covered in the chapter?

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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

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

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

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

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Please register to view this section

FAQs (Frequently Asked Questions)

1. What topics are covered in the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3?

2. What is the number of questions in the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3?

3. Where can a student find the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3?

4. How do the NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3 help students score higher in CBSE board exams?

5. Do NCERT Solutions Class 12 Maths Chapter 11 Exercise 11.3 help students understand the concepts covered in the chapter?

Share

NCERT Solutions Related Links

NCERT Solutions

Solved Board Papers

CBSE Class

ICSE Class

Sample Paper

Exam Weightage

Board Paper Solution 2020

Test Preparation

Other Services

About Us

Terms & Conditions