The distance of the point ( α , β , γ ) from the y-axis is (A) β (B) γ (C) β 2 + γ 2 (D) α 2 + γ 2

This is a Objective Type Questions as classified in NCERT Exemplar The given point is (α, β, γ) Any point on y−axis=(0, β, 0) ∴Required distance=(α−0)2+(β−β)2+(δ−0)2 = α 2 + γ 2 H e n c e , t h e c o r r e c t o p t i o n i s ( d ) .

The distance of the plane from the origin is: (A) 1 (B) 7 (C) 1 7 (D) None of these

This is a Objective Type Questions as classified in NCERT Exemplar G i v e n t h a t r → . ( 2 7 i ^ + 3 7 j ^ − 6 7 k ^ ) = 1 So, the distance of the given plane from the origin is = | − 1 ( 2 7 ) 2 + ( 3 7 ) 2 + ( − 6 7 ) 2 | = | − 1 4 4 9 + 9 4 9 + 3 6 4 9 | = 1 1 = 1 H e n c e , t h e c o r r e c t o p t i o n i s ( a ) .

The angle between the line r → = ( 5 i ^ + 3 j ^ + 7 k ^ ) + λ ( 2 i ^ − j ^ + k ^ ) and the plane r → = 3 i ^ − 4 j ^ − 4 k ^ + 5 = 0 is s i n − 1 ( 5 / 2 √ 9 1 ) . .

This is a True or False Type Questions as classified in NCERT Exemplar E q u a t i o n o f l i n e i s r → = ( 5 i ^ − j ^ − 4 k ^ ) + λ ( 2 i ^ − j ^ + k ^ ) a n d t h e e q u a t i o n o f p l a n e i s r → . ( 3 i ^ − 4 j ^ − k ^ ) + 5 = 0 H e r e , b 1 → = 2 i ^ − j ^ + k ^ a n d n 2 → = 3 i ^ − 4 j ^ − k ^ ∴ s i n θ = b 1 → n 2 → | b 1 → | | n 2 → | ⇒ s i n θ = ( 2 i ^ − j ^ + k ^ ) . ( 3 i ^ − 4 j ^ − k ^ ) 4 + 4 + 1 9 + 1 6 + 1 = 6 + 4 − 1 6 2 6 = 9 6 2 6 ⇒ s i n θ = 9 2 3 9 w h i c h i s f a l s e . H e n c e , t h e g i v e n s t a t e m e n t i s F a l s e .

Find the length and the foot of the perpendicular from the point ( 1 , 3 2 , 2 ) to the plane 2 x − 2 y + 4 z + 5 = 0 .

This is a long answer type question as classified in NCERT Exemplar Given plane is 2x−2y+4z+5=0 and the given points is(1, 32, 2) D ' r a t i o s o f t h e n o r m a l t o t h e p l a n e a r e 2 , − 2 , 4 So, the? equation of line passing through (1, 32, 2) and whose d'ratios are equal to the D ' r a t i o s o f t h e n o r m a l t o t h e p l a n e i . e . , 2 , − 2 , 4 i s x − 1 2 = y − 3 2 − 2 = z − 2 4 = λ ∴Any point in the plane is 2λ+1, −2λ+32, 4λ+2 Since, the point lies in the plane, then 2 ( 2 λ + 1 ) − 2 ( − 2 λ + 3 2 ) + 4 ( 4 λ + 2 ) + 5 = 0 ⇒ 4 λ + 2 + 4 λ − 3 + 1 6 λ + 8 + 5 = 0 ⇒ 2 4 λ + 1 2 = 0 ∴ λ = − 1 2 So, the coordinates of the point in the plane are 2 ( − 1 2 ) + 1 , − 2 ( − 1 2 ) + 3 2 , 4 ( − 1 2 ) + 2 i . e . , 0 , 5 2 , 0 H e n c e , 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 i s ( 0 , 5 2 , 0 ) a n d t h e r e q u i r e d l e n g t h = ( 1 − 0 ) 2 + ( 3 2 − 5 2 ) 2 + ( 2 − 0 ) 2 = 1 + 1 + 4 = 6 u n i t s

The direction cosines of the vector ( 2 i ^ + 2 j ^ − k ^ ) are __________.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar L e t a → = 2 i ^ + 2 j ^ − k ^ d i r e c t i o n r a t i o s o f a → a r e 2 , 2 , − 1 ∴the direction cosines are 24+4+1, 24+4+1, −14+4+1 ⇒ 2 3 , 2 3 , − 1 3 Hence, the direction cosines of the given vector are 23, 23, −13.

The vector equation of the line x − 5 3 = y + 4 7 = z − 6 2 is __________.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar T h e g i v e n e q u a t i o n i s x − 5 3 = y + 4 7 = z − 6 2 H e r e , a → = ( 5 i ^ − 4 j ^ + 6 k ^ ) a n d b → = ( 3 i ^ + 7 j ^ + 2 k ^ ) E q u a t i o n o f t h e l i n e i s r → = a → + b → λ H e n c e , t h e v e c t o r e q u a t i o n o f t h e g i v e n l i n e i s r → = ( 5 i ^ − 4 j ^ + 6 k ^ ) + λ ( 3 i ^ + 7 j ^ + 2 k ^ )

Find the position vector of a point A in space such that O A → is inclined at 60º to OX and at 45° to OY and | O A → | = 10 units.

This is a short answer type question as classified in NCERT Exemplar L e t α = 6 0 0 , β = 4 5 0 a n d t h e a n g l e i n c l i n e d t o O Z a x i s b e γ . W e k n o w t h a t c o s 2 α + c o s 2 β + c o s 2 γ = 1 ⇒ c o s 2 6 0 0 + c o s 2 4 5 0 + c o s 2 γ = 1 ⇒ ( 1 2 ) 2 + ( 1 2 ) 2 + c o s 2 γ = 1 ⇒ 1 4 + 1 2 + c o s 2 γ = 1 ⇒ 3 4 + c o s 2 γ = 1 ⇒ c o s 2 γ = 1 − 3 4 = 1 4 ∴ c o s γ = ± 1 2 ⇒ c o s γ = 1 2 ( Rejecting cosγ=−12, since γ<900 ) ∴ O A → = | O A → | ( 1 2 i ^ + 1 2 j ^ + 1 2 k ^ ) = 1 0 ( 1 2 i ^ + 1 2 j ^ + 1 2 k ^ ) = 5 i ^ + 5 2 j ^ + 5 k ^ H e n c e , t h e p o s i t i o n v e c t o r o f A i s ( 5 i ^ + 5 2 j ^ + 5 k ^ ) .

Show that the lines x − 1 2 = y − 2 3 = z − 3 4 and x − 4 5 = y − 1 2 = z intersect. Also, find their point of intersection.

This is a short answer type question as classified in NCERT Exemplar T h e g i v e n e q u a t i o n s a r e x − 1 2 = y − 2 3 = z − 3 4 a n d x − 4 5 = y − 1 2 = z L e t x − 1 2 = y − 2 3 = z − 3 4 = λ ∴ x = 2 λ + 1 , y = 3 λ + 2 a n d z = 4 λ + 3 a n d x − 4 5 = y − 1 2 = z 1 = μ ∴ x = 5 μ + 4 , y = 2 μ + 1 a n d z = μ If the two lines intersect each other at one point, t h e n 2 λ + 1 = 5 μ + 4 ⇒ 2 λ − 5 μ = 3 … ( i ) 3 λ + 2 = 2 μ + 1 ⇒ 3 λ − 2 μ = − 1 … ( i i ) a n d 4 λ + 3 = μ ⇒ 4 λ − μ = − 3 … ( i i i ) S o l v i n g e q n . ( i ) a n d ( i i ) w e g e t 2 λ − 5 μ = 3 ( M u l t i p l y b y 3 ) 3 λ − 2 μ = − 1 ( M u l t i p l y b y 2 ) ⇒ 6 λ − 1 5 μ = 9 6 λ − 4 μ = − 2 ( − ) ( + ) ( + ) _ − 1 1 μ = 1 1 ∴ μ = − 1 P u t t i n g t h e v a l u e o f μ i n e q n . ( i ) w e g e t , 2 λ − 5 ( − 1 ) = 3 ⇒ 2 λ + 5 = 3 ⇒ 2 λ = − 2 ∴ λ = − 1 N o w , p u t t i n g t h e v a l u e o f λ μ i n e q n . ( i i i ) t h e n , 4 ( − 1 ) − ( − 1 ) = − 3 − 4 + 1 = − 3 − 3 = − 3 ( s a t i s f i e d ) ∴ Coordinates of the point of intersection are x = 5 ( − 1 ) + 4 = − 5 + 4 = − 1 y = 2 ( − 1 ) + 1 = − 2 + 1 = − 1 z = − 1 Hence, the given lines intersect each other at (−1,−1,−1).

Find the vector equation of the line which is parallel to the vector 3 i ^ − 2 j ^ + 6 k ^ and which passes through the point (1, –2, 3).

This is a short answer type question as classified in NCERT Exemplar W e k n o w t h a t t h e e q u a t i o n o f l i n e i s r → = a → + b → λ H e r e , a → = i ^ − 2 j ^ + 3 k ^ a n d b → = 3 i ^ − 2 j ^ + 6 k ^ ∴ E q u a t i o n o f l i n e i s r → = ( i ^ − 2 j ^ + 3 k ^ ) + λ ( 3 i ^ − 2 j ^ + 6 k ^ ) ⇒ ( x i ^ + y j ^ + z k ^ ) = ( i ^ − 2 j ^ + 3 k ^ ) + λ ( 3 i ^ − 2 j ^ + 6 k ^ ) ⇒ ( x i ^ + y j ^ + z k ^ ) − ( i ^ − 2 j ^ + 3 k ^ ) = λ ( 3 i ^ − 2 j ^ + 6 k ^ ) ⇒ ( x − 1 ) i ^ + ( y + 2 ) j ^ + ( z − 3 ) k ^ = λ ( 3 i ^ − 2 j ^ + 6 k ^ ) H e n c e , t h e r e q u i r e d e q u a t i o n i s ( x − 1 ) i ^ + ( y + 2 ) j ^ + ( z − 3 ) k ^ = λ ( 3 i ^ − 2 j ^ + 6 k ^ ) .

Maths NCERT Exemplar Solutions Class 12th Chapter Eleven: Overview, Questions, Preparation

Updated on Jul 15, 2025 10:48 IST

By Payal Gupta, Retainer

Table of content

Three-Dimensional Geometry Long Answer Type Questions
Three-Dimensional Geometry Short Answer Type Questions
Three-Dimensional Geometry Objective Type Questions
Three-Dimensional Geometry Fill in the blanks Type Questions
Three-Dimensional Geometry True or False Type Questions

Three-Dimensional Geometry Long Answer Type Questions

1. Find the foot of the perpendicular from the point (2, 3, –8) to the line $\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .$ Also, find the perpendicular distance from the given point to the line.

Ans.

\begin{array}{l} G i v e n t h a t \frac{4 - x}{2} + \frac{y}{6} + \frac{1 - z}{3} i s t h e e q u a t i o n o f l i n e \\ \Rightarrow \frac{x - 4}{- 2} + \frac{y}{6} + \frac{z - 1}{- 3} = λ \\ ∴ C o o r d i n a t e s o f a n y point Q o n t h e l i n e a r e \\ x = - 2 λ + 4, y = 6 λ a n d z = - 3 λ + 1 \\ a n d t h e g i v e n point i s P (2, 3, - 8) \\ D i r e c t i o n r a t i o s o f P Q a r e - 2 λ + 4 - 2, 6 λ - 3, - 3 λ + 1 + 8 \\ a n d t h e D' r a t i o s o f t h e g i v e n l i n e a r e - 2, 6, - 3 . \\ I f P Q ⊥ l i n e \\ t h e n - 2 (- 2 λ + 2) + 6 (6 λ - 3) - 3 (- 3 λ + 9) = 0 \\ \Rightarrow 4 λ - 4 + 36 λ - 18 + 9 λ - 27 = 0 \\ \Rightarrow 49 λ - 49 = 0 \Rightarrow λ = 1 \\ ∴ 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 i s - 2 (1) + 4, 6 (1), - 3 (1) + 1 \\ i . e ., 2, 6, - 2 \\ N o w, distance P Q = \sqrt[]{{(2 - 2)}^{2} + {(3 - 6)}^{2} + {(- 8 + 2)}^{2}} \\ = \sqrt[]{9 + 36} = \sqrt[]{45} = 3 \sqrt[]{5} \\ H e n c e, t h e r e q u i r e d 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 2, 6, - 2 a n d t h e \\ r e q u i r e d distance 3 \sqrt[]{5} u n i t s . \end{array}

2. Find the distance of the point (2, 4, –1) from the line $\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{– 9} .$

Ans. $\begin{array}{l} T h e g i v e n e q u a t i o n o f l i n e i s \\ \frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{– 9} = λ a n d a n y point P (2, 4, - 1) \\ L e t Q b e a n y point o n t h e g i v e n l i n e \\ ∴ C o o r d i n a t e s o f Q a r e x = λ - 5, y = 4 λ - 3 a n d z = - 9 λ + 6 \\ a n d t h e g i v e n point i s P (2, 3, - 8) \\ D i r e c t i o n r a t i o s o f P Q a r e λ - 5 - 2, 4 λ - 3 - 4, - 9 λ + 6 + 1 \\ i . e ., λ - 7, 4 λ - 7, - 9 λ + 7 \\ a n d t h e D' r a t i o s o f t h e g i v e n l i n e a r e 1, 4, - 9 . \\ I f P Q ⊥ l i n e \\ t h e n 1 (λ - 7) + 4 (4 λ - 7) - 9 (- 9 λ + 7) = 0 \\ \Rightarrow λ - 7 + 16 λ - 28 + 81 λ - 63 = 0 \\ \Rightarrow 98 λ - 98 = 0 \Rightarrow λ = 1 \\ S o, C o o r d i n a t e s o f Q a r e 1 - 5, 4 \times 1 - 3, - 9 \times 1 + 6 \\ i . e ., - 4, 1, - 3 \\ N o w, distance P Q = \sqrt[]{{(- 4 - 2)}^{2} + {(1 - 4)}^{2} + {(- 3 + 1)}^{2}} \\ = \sqrt[]{{(- 6)}^{2} + {(- 3)}^{2} + {(- 2)}^{2}} = \sqrt[]{36 + 9 + 4} = \sqrt[]{49} = 7 \\ H e n c e, t h e r e q u i r e d distance 7 u n i t s . \end{array}$

Commonly asked questions

Find the foot of the perpendicular from the point (2, 3, –8) to the line $\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .$ Also, find the perpendicular distance from the given point to the line.

Find the distance of the point (2, 4, –1) from the line $\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{– 9} .$

Find the length and the foot of the perpendicular from the point ( $1, \frac{3}{2}, 2$ ) to the plane $2 x - 2 y + 4 z + 5 = 0$ .

Three-Dimensional Geometry Short Answer Type Questions

1. Find the position vector of a point A in space such that $\vec{O A}$ is inclined at 60º to OX and at 45° to OY and $| \vec{O A} |$ = 10 units.

Ans.

\begin{array}{l} L e t α = 6 0^{0}, β = 4 5^{0} a n d t h e a n g l e i n c l i n e d t o O Z a x i s b e γ . \\ W e k n o w t h a t c o s^{2} α + c o s^{2} β + c o s^{2} γ = 1 \\ \Rightarrow c o s^{2} 6 0^{0} + c o s^{2} 4 5^{0} + c o s^{2} γ = 1 \\ \Rightarrow {(\frac{1}{2})}^{2} + {(\frac{1}{\sqrt[]{2}})}^{2} + c o s^{2} γ = 1 \Rightarrow \frac{1}{4} + \frac{1}{2} + c o s^{2} γ = 1 \\ \Rightarrow \frac{3}{4} + c o s^{2} γ = 1 \Rightarrow c o s^{2} γ = 1 - \frac{3}{4} = \frac{1}{4} \\ ∴ c o s γ = \pm \frac{1}{2} \Rightarrow c o s γ = \frac{1}{2} (Rejecting c o s γ = - \frac{1}{2}, since γ < 9 0^{0}) \\ ∴ \vec{O A} = | \vec{O A} | (\frac{1}{2} \hat{i} + \frac{1}{\sqrt[]{2}} \hat{j} + \frac{1}{2} \hat{k}) = 10 (\frac{1}{2} \hat{i} + \frac{1}{\sqrt[]{2}} \hat{j} + \frac{1}{2} \hat{k}) \\ = 5 \hat{i} + 5 \sqrt[]{2} \hat{j} + 5 \hat{k} \\ H e n c e, t h e p o s i t i o n v e c t o r o f A i s (5 \hat{i} + 5 \sqrt[]{2} \hat{j} + 5 \hat{k}) . \end{array}

2. Find the vector equation of the line which is parallel to the vector $3 \hat{i} - 2 \hat{j} + 6 \hat{k}$ and which passes through the point (1, –2, 3).

Ans. $\begin{array}{l} W e k n o w t h a t t h e e q u a t i o n o f l i n e i s \\ \vec{r} = \vec{a} + \vec{b} λ \\ H e r e, \vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k} a n d \vec{b} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k} \\ ∴ E q u a t i o n o f l i n e i s \vec{r} = (\hat{i} - 2 \hat{j} + 3 \hat{k}) + λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) \\ \Rightarrow (x \hat{i} + y \hat{j} + z \hat{k}) = (\hat{i} - 2 \hat{j} + 3 \hat{k}) + λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) \\ \Rightarrow (x \hat{i} + y \hat{j} + z \hat{k}) - (\hat{i} - 2 \hat{j} + 3 \hat{k}) = λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) \\ \Rightarrow (x - 1) \hat{i} + (y + 2) \hat{j} + (z - 3) \hat{k} = λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s (x - 1) \hat{i} + (y + 2) \hat{j} + (z - 3) \hat{k} = λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) . \end{array}$

Commonly asked questions

Find the position vector of a point A in space such that $\vec{O A}$ is inclined at 60º to OX and at 45° to OY and $| \vec{O A} |$ = 10 units.

Show that the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = z$ intersect. Also, find their point of intersection.

Find the vector equation of the line which is parallel to the vector $3 \hat{i} - 2 \hat{j} + 6 \hat{k}$ and which passes through the point (1, –2, 3).

Three-Dimensional Geometry Objective Type Questions

1. The distance of the point $(α, β, γ)$ from the y-axis is

(A) $β$

(B) $γ$

(C) $\sqrt[]{β^{2} + γ^{2}}$

(D) $α^{2} + γ^{2}$

Ans. $\begin{array}{l} T h e g i v e n point i s (α, β, γ) \\ A n y point o n y - a x i s = (0, β, 0) \\ ∴ Required distance = \sqrt[]{{(α - 0)}^{2} + {(β - β)}^{2} + {(δ - 0)}^{2}} \\ = \sqrt[]{α^{2} + γ^{2}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

2. If the direction cosines of a line are $k, k, k$ , then:

(A) $k > 0$

(B) $0 < k < 1$

(C) $k = 1$

(D) $k = \frac{1}{\sqrt[]{3}}$ or - $\frac{1}{\sqrt[]{3}}$

Ans. $\begin{array}{l} I f l, m, n a r e t h e d i r e c t i o n c o s i n e s o f a l i n e, t h e n \\ l^{2} + m^{2} + n^{2} = 1 \\ S o, k^{2} + k^{2} + k^{2} = 1 \\ \Rightarrow 3 k^{2} = 1 \Rightarrow k = \pm \frac{1}{\sqrt[]{3}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Commonly asked questions

The distance of the point $(α, β, γ)$ from the y-axis is

(A) $β$

(B) $γ$

(C) $\sqrt[]{β^{2} + γ^{2}}$

(D) $α^{2} + γ^{2}$

If the direction cosines of a line are $k, k, k$ , then:

(A) $k > 0$

(B) $0 < k < 1$

(C) $k = 1$

(D) $k = \frac{1}{\sqrt[]{3}}$ or - $\frac{1}{\sqrt[]{3}}$

The distance of the plane from the origin is:

(A) 1

(B) 7

(C) $\frac{1}{7}$

(D) None of these

Three-Dimensional Geometry Fill in the blanks Type Questions

1. A plane passes through the points (2, 0, 0), (0, 3, 0), and (0, 0, 4). The equation of the plane is __________.

Ans. $\begin{array}{l} G i v e n points a r e (2, 0, 0), (0, 3, 0) a n d (0, 0, 4) . \\ S o, t h e intercepts c u t s b y t h e p l a n e o n t h e a x e s a r e 2, 3, 4 \\ E q u a t i o n o f t h e p l a n e (intercept f o r m) i s \\ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \Rightarrow \frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1 \\ H e n c e, t h e e q u a t i o n o f t h e p l a n e i s \frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1 \end{array}$

2. The direction cosines of the vector $(2 \hat{i} + 2 \hat{j} - \hat{k})$ are __________.

Ans. $\begin{array}{l} L e t \vec{a} = 2 \hat{i} + 2 \hat{j} - \hat{k} \\ d i r e c t i o n r a t i o s o f \vec{a} a r e 2, 2, - 1 \\ ∴ t h e d i r e c t i o n cosines a r e \frac{2}{\sqrt[]{4 + 4 + 1}}, \frac{2}{\sqrt[]{4 + 4 + 1}}, \frac{- 1}{\sqrt[]{4 + 4 + 1}} \\ \Rightarrow \frac{2}{3}, \frac{2}{3}, \frac{- 1}{3} \\ H e n c e, t h e d i r e c t i o n cosines o f t h e g i v e n v e c t o r a r e \frac{2}{3}, \frac{2}{3}, \frac{- 1}{3} . \end{array}$

Commonly asked questions

A plane passes through the points (2, 0, 0), (0, 3, 0), and (0, 0, 4). The equation of the plane is __________.

The direction cosines of the vector $(2 \hat{i} + 2 \hat{j} - \hat{k})$ are __________.

The vector equation of the line $\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2}$ is __________.

Three-Dimensional Geometry True or False Type Questions

1. The unit vector normal to the plane $x + 2 y + 3 z - 6 = 0$ is $1 / \sqrt 14 \hat{i} + 2 / \sqrt 14 \hat{j} + 3 / \sqrt 14 \hat{k} .$

Ans. $\begin{array}{l} G i v e n p l a n e i s x + 2 y + 3 z - 6 = 0 \\ V e c t o r n o r m a l t o t h e p l a n e \vec{n} = \hat{i} + 2 \hat{j} + 3 \hat{k} \\ ∴ \hat{n} = \frac{\vec{n}}{| \vec{n} |} = \frac{\hat{i} + 2 \hat{j} + 3 \hat{k}}{\sqrt[]{{(1)}^{2} + {(2)}^{2} + {(3)}^{2}}} = \frac{1}{\sqrt[]{14}} \hat{i} + \frac{2}{\sqrt[]{14}} \hat{j} + \frac{3}{\sqrt[]{14}} \hat{k} \\ H e n c e, t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

2. The intercepts made by the plane $2 x - 3 y + 5 z + 4 = 0$ on the coordinate axis are $- 2, \frac{4}{3}, - \frac{4}{5}$ .

Ans. $\begin{array}{l} E q u a t i o n o f t h e p l a n e i s 2 x - 3 y + 5 z + 4 = 0 \\ \Rightarrow 2 x - 3 y + 5 z = - 4 \\ \Rightarrow \frac{2}{- 4} x - \frac{3}{- 4} y + \frac{5}{- 4} z = 1 \\ \Rightarrow \frac{x}{- 2} - \frac{y}{4 / 3} + \frac{z}{- 4 / 5} = 1 \\ S o, t h e r e q u i r e d intercepts a r e - 2, \frac{4}{3} a n d \frac{- 4}{5} \\ H e n c e, t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

Commonly asked questions

The unit vector normal to the plane $x + 2 y + 3 z - 6 = 0$ is $1 / \sqrt 14 \hat{i} + 2 / \sqrt 14 \hat{j} + 3 / \sqrt 14 \hat{k} .$

The intercepts made by the plane $2 x - 3 y + 5 z + 4 = 0$ on the coordinate axis are $- 2, \frac{4}{3}, - \frac{4}{5}$ .

The angle between the line $\vec{r} = (5 \hat{i} + 3 \hat{j} + 7 \hat{k})$ + $λ (2 \hat{i} - \hat{j} + \hat{k})$ and the plane $\vec{r} = \hat{3 i} - 4 \hat{j} - 4 \hat{k}$ + 5 = 0 is $s i n^{- 1} (5 / 2 \sqrt 91)$ . .

Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exam

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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven: Overview, Questions, Preparation

Three-Dimensional Geometry Long Answer Type Questions

Commonly asked questions

Three-Dimensional Geometry Short Answer Type Questions

Commonly asked questions

Three-Dimensional Geometry Objective Type Questions

Commonly asked questions

Three-Dimensional Geometry Fill in the blanks Type Questions

Commonly asked questions

Three-Dimensional Geometry True or False Type Questions

Commonly asked questions

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