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Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The area of the quadrilateral ABCD, where A(0, 4, 1), B(2, 3, –1), C(4, 5, 0) and D(2, 6, 2), is equal to:

(A) 9 sq. units

(B) 18 sq. units

(C) 27 sq. units

(D) 81 sq. units

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Payal Gupta

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n points a r e A (0, 4, 1), B (2, 3, - 1), C (4, 5, 0) a n d D (2, 6, 2) \\ d' r a t i o s o f A B = 2, - 1, - 2 \\ a n d d' r a t i o s o f D C = 2, - 1, - 2 \\ ∴ A B ? D C \\ S o, ? A B C D i s a parallelogram . \\ \vec{A B} = 2 \hat{i} - \hat{j} + 2 \hat{k} a n d \vec{A D} = 2 \hat{i} + 2 \hat{j} + \hat{k} \\ ∴ A r e a o f parallelogram A B C D = | \vec{A B} * \vec{A D} | \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 1 & - 2 \\ 2 & 2 & 1 \end{matrix} | \\ = \hat{i} (- 1 + 4) - \hat{j} (2 + 4) + \hat{k} (4 + 2) = 3 \hat{i} - 6 \hat{j} + 6 \hat{k} \\ = \sqrt[]{{(3)}^{2} + {(- 6)}^{2} + {(6)}^{2}} = \sqrt[]{9 + 36 + 36} = \sqrt[]{81} = 9 S q u n i t s \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The reflection of the point $(α, β, γ)$ in the xy-plane is

(A) $(α, β, 0)$

(B) $(0, 0, γ)$

(C) $(- α, - β, γ)$

(D) $(α, β, - γ)$

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Payal Gupta

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} Reflextion o f point (α, β, γ) i n x y - p l a n e i s (α, β, - γ) \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The sine of the angle between the straight line $\frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ and the plane $2 x - 2 y + z = 5$ is:

(A) $\frac{10}{6 \sqrt[]{5}}$

(B) $\frac{4}{5 \sqrt[]{2}}$

(C) $\frac{2}{3 \sqrt[]{5}}$

(D) $\frac{\sqrt 2}{10}$

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Payal Gupta

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t l : \frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5} \\ a n d P : 2 x - 2 y + z = 5 \\ d' r a t i o s o f t h e l i n e a r e 3, 4, 5 \\ a n d d' r a t i o s o f t h e n o r m a l t o t h e p l a n e a r e 2, - 2, 1 \\ ∴ s i n θ = \frac{3 (2) + 4 (- 2) + 5 (1)}{\sqrt[]{9 + 16 + 25} . \sqrt[]{4 + 4 + 1}} \\ \Rightarrow s i n θ = \frac{6 - 8 + 5}{\sqrt[]{50} . 3} = \frac{3}{5 \sqrt[]{2}} = \frac{\sqrt[]{2}}{10} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The distance of the plane from the origin is:

(A) 1

(B) 7

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

(D) None of these

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Payal Gupta

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t \vec{r} . (\frac{2}{7} \hat{i} + \frac{3}{7} \hat{j} - \frac{6}{7} \hat{k}) = 1 \\ S o, t h e distance o f t h e g i v e n p l a n e f r o m t h e o r i g i n i s \\ = | \frac{- 1}{\sqrt[]{{(\frac{2}{7})}^{2} + {(\frac{3}{7})}^{2} + {(\frac{- 6}{7})}^{2}}} | = | \frac{- 1}{\sqrt[]{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}}} | = \frac{1}{1} = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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 - $\frac{1}{\sqrt[]{3}}$

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Payal Gupta

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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

(A) $β$

(B) $γ$

(C) $\sqrt[]{β^{2} + γ^{2}}$

(D) $α^{2} + γ^{2}$

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Payal Gupta

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

If $l_{1}, m_{1}, n_{1}$ ; $l_{2}, m_{2}, n_{2}$ , and $l_{3}, m_{3}, n_{3}$ are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to $l_{1} + l_{2} + l_{3}$ , $m_{1} + m_{2} + m_{3}$ , $n_{1} + n_{2} + n_{3}$ and makes equal angles with them.

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} L e t \vec{a}, \vec{b}, \vec{c} a n d \vec{d} a r e s u c h t h a t \\ \vec{a} = l_{1} \hat{i} + m_{1} \hat{j} + n_{1} \hat{k} \\ \vec{b} = l_{2} \hat{i} + m_{2} \hat{j} + n_{2} \hat{k} \\ \vec{c} = l_{3} \hat{i} + m_{3} \hat{j} + n_{3} \hat{k} \\ a n d \vec{d} = (l_{1} + l_{2} + l_{3}) \hat{i} + (m_{1} + m_{2} + m_{3}) \hat{j} + (n_{1} + n_{2} + n_{3}) \hat{k} \\ Since t h e g i v e n d' cosines a r e m u t u a l l y p e r p e n d i c u l a r t h e n \\ l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0 \\ l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3} = 0 \\ l_{1} l_{3} + m_{1} m_{3} + n_{1} n_{3} = 0 \\ L e t α, β a n d γ b e t h e a n g l e s b e t w e e n \vec{a} a n d \vec{d}, \vec{b} a n d \vec{d}, \vec{c} a n d \vec{d} r e s p e c t i v e l y . \\ ∴ c o s α = l_{1} (l_{1} + l_{2} + l_{3}) + m_{1} (m_{1} + m_{2} + m_{3}) + n_{1} (n_{1} + n_{2} + n_{3}) \\ = l_{1}^{2} + l_{1} l_{2} + l_{1} l_{3} + m_{1}^{2} + m_{1} m_{2} + m_{1} m_{3} + n_{1}^{2} + n_{1} n_{2} + n_{1} n_{3} \\ = (l_{1}^{2} + m_{1}^{2} + n_{1}^{2}) + (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}) + (l_{1} l_{3} + m_{1} m_{3} + n_{1} n_{3}) \\ = 1 + 0 + 0 = 1 \\ ∴ c o s β = l_{2} (l_{1} + l_{2} + l_{3}) + m_{2} (m_{1} + m_{2} + m_{3}) + n_{2} (n_{1} + n_{2} + n_{3}) \\ = l_{1} l_{2} + l_{2}^{2} + l_{2} l_{3} + m_{1} m_{2} + m_{2}^{2} + m_{2} m_{3} + n_{1} n_{2} + n_{2}^{2} + n_{2} n_{3} \\ = (l_{2}^{2} + m_{2}^{2} + n_{2}^{2}) + (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}) + (l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3}) \\ = 1 + 0 + 0 = 1 \\ S i m i l a r l y, \\ ∴ c o s γ = l_{3} (l_{1} + l_{2} + l_{3}) + m_{3} (m_{1} + m_{2} + m_{3}) + n_{3} (n_{1} + n_{2} + n_{3}) \\ = l_{1} l_{3} + l_{2} l_{3} + l_{3}^{2} + m_{1} m_{3} + m_{2} m_{3} + m_{3}^{2} + n_{1} n_{3} + n_{2} n_{3} + n_{3}^{2} \\ = (l_{3}^{2} + m_{3}^{2} + n_{3}^{2}) + (l_{1} l_{3} + m_{1} m_{3} + n_{1} n_{3}) + (l_{2} l_{3} + m_{2} m_{3} + n_{2} n_{3}) \\ = 1 + 0 + 0 = 1 \\ ∴ c o s α = c o s β = c o s γ = 1 \Rightarrow α = β = γ w h i c h i s t h e r e q u i r e d r e s u l t . \end{array}$

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Show that the straight lines whose direction cosines are given by $2 l + 2 m - n = 0 and m n + n l + l m = 0$ are at right angles.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t 2 l + 2 m - n = 0 \dots (i) \\ a n d m n + n l + l m = 0 \dots (i i) \\ Eliminating m f r o m e q . (i) a n d (i i) w e g e t, \\ m = \frac{n - 2 l}{2} [F r o m (i)] \\ \Rightarrow (\frac{n - 2 l}{2}) n + n l + l (\frac{n - 2 l}{2}) = 0 \\ \Rightarrow \frac{n^{2} - 2 n l + 2 n l + n l - 2 l^{2}}{2} = 0 \\ \Rightarrow n^{2} + n l - 2 l^{2} = 0 \\ \Rightarrow n^{2} + 2 n l - n l - 2 l^{2} = 0 \\ \Rightarrow n (n + 2 l) - l (n + 2 l) = 0 \\ \Rightarrow (n - l) (n + 2 l) = 0 \\ \Rightarrow n = - 2 l a n d n = l \\ ∴ m = \frac{- 2 l - 2 l}{2}, m = \frac{l - 2 l}{2} \\ \Rightarrow m = - 2 l, m = \frac{- l}{2} \\ T h e r e f o r e, t h e d i r e c t i o n r a t i o s a r e p r o p o r t i o n a l t o l, - 2 l, - 2 l a n d l, \frac{- l}{2}, l . \\ \Rightarrow 1, - 2, - 2 a n d 2, - 1, 2 \\ I f t h e t w o l i n e s a r e p e r p e n d i c u l a r t o e a c h o t h e r t h e n \\ 1 (2) - 2 (- 1) - 2 * 2 = 0 \\ 2 + 2 - 4 = 0 \\ 0 = 0 \\ H e n c e, t h e t w o l i n e s a r e p e r p e n d i c u l a r . \end{array}$

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

AB = 3 $\hat{i}$ - $\hat{j}$ + $\hat{k}$ and $C D$ = -3 $\hat{i}$ + 2 $\hat{j}$ + 4 $\hat{k}$ are two vectors. The position vectors of the points A and C are 6 $\hat{i}$ + 7 $\hat{j}$ + 4 $\hat{k}$ and -9 $\hat{i}$ + 2 $\hat{j}$ + $\hat{k,}$ respectively. Find the position vector of a point P on the line AB and a point Q on the line CD such that $\vec{P Q}$ is perpendicular to both $\vec{A B}$ and $\vec{C D}$ both.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} P o s i t i o n v e c t o r o f A i s 6 \hat{i} + 7 \hat{j} + 4 \hat{k} a n d \vec{A B} = 3 \hat{i} - \hat{j} + \hat{k} \\ S o, e q u a t i o n o f a n y l i n e passing t h r o u g h A a n d p a r a l l e l t o \vec{A B} \\ \vec{r} = (6 \hat{i} + 7 \hat{j} + 4 \hat{k}) + λ (3 \hat{i} - \hat{j} + \hat{k}) \dots (i) \\ N o w a n y point P o n \vec{A B} = (6 + 3 λ, 7 - λ, 4 + λ) \\ S i m i l a r l y, p o s i t i o n v e c t o r o f C i s - 9 \hat{j} + 2 \hat{k} a n d \vec{C D} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k} \\ S o, e q u a t i o n o f a n y l i n e passing t h r o u g h C a n d p a r a l l e l t o \vec{C D} i s \\ \vec{r} = (- 9 \hat{j} + 2 \hat{k}) + μ (- 3 \hat{i} + 2 \hat{j} + 4 \hat{k}) \dots (i i) \\ A n y point Q o n \vec{C D} = (- 3 μ, - 9 + 2 μ, 2 + 4 μ) \\ d' r a t i o s o f \vec{P Q} a r e \\ (6 + 3 λ + 3 μ, 7 - λ + 9 - 2 μ, 4 + λ - 2 - 4 μ) \\ \Rightarrow (6 + 3 λ + 3 μ), (16 - λ - 2 μ), (2 + λ - 4 μ) \\ N o w \vec{P Q} i s ⊥ t o e q . (i), t h e n \\ 3 (6 + 3 λ + 3 μ) - 1 (16 - λ - 2 μ) + 1 (2 + λ - 4 μ) = 0 \\ \Rightarrow 18 + 9 λ + 9 μ - 16 + λ + 2 μ + 2 + λ - 4 μ = 0 \\ \Rightarrow 11 λ + 7 μ + 4 = 0 \dots (i i i) \\ \vec{P Q} i s ⊥ t o e q . (i i), t h e n \\ - 3 (6 + 3 λ + 3 μ) + 2 (16 - λ - 2 μ) + 4 (2 + λ - 4 μ) = 0 \\ \Rightarrow - 18 - 9 λ - 9 μ + 32 - 2 λ - 4 μ + 8 + 4 λ - 16 μ = 0 \\ \Rightarrow - 7 λ - 29 μ + 22 = 0 \\ \Rightarrow 7 λ + 29 μ - 22 = 0 \dots (i v) \\ S o l v i n g e q n . (i i i) a n d (i v) w e g e t \\ \Rightarrow 77 λ + 49 μ + 28 = 0 \\ 77 λ + 319 μ - 242 = 0 \\ \underline{(-) (-) (+)} \\ - 270 μ + 270 = 0 ∴ 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 v) w e g e t, \\ 7 λ + 29 - 22 = 0 \Rightarrow λ = - 1 \\ ∴ P o s i t i o n v e c t o r o f P = [6 + 3 (- 1), 7 + 1, 4 - 1] = (3, 8, 3) \\ a n d P o s i t i o n v e c t o r o f Q = [- 3 (1), - 9 + 2 (1), 2 + 4 (1)] = (- 3, - 7, 6) \\ H e n c e, t h e p o s i t i o n v e c t o r o f \\ P = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} a n d Q = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} \end{array}$

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Show that the points $(\hat{i} - \hat{j} + \hat{k})$ and 3 $(\hat{i} + \hat{j} + \hat{k})$ are equidistant from the plane and lie on opposite sides of it.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n points a r e P (\hat{i} - \hat{j} + 3 \hat{k}) a n d Q (3 \hat{i} + 3 \hat{j} + 3 \hat{k}) a n d t h e p l a n e \vec{r} . (5 \hat{i} + 2 \hat{j} - 7 \hat{k}) + 9 = 0 \\ P e r p e n d i c u l a r distance o f P (\hat{i} - \hat{j} + 3 \hat{k}) f r o m t h e p l a n e \\ \vec{r} . (5 \hat{i} + 2 \hat{j} - 7 \hat{k}) + 9 = | \frac{(\hat{i} - \hat{j} + 3 \hat{k}) . (5 \hat{i} + 2 \hat{j} - 7 \hat{k}) + 9}{\sqrt[]{{(5)}^{2} + {(2)}^{2} + {(- 7)}^{2}}} | \\ = | \frac{5 - 2 - 21 + 9}{\sqrt[]{25 + 4 + 49}} | = | \frac{- 9}{\sqrt[]{78}} | \\ a n d p e r p e n d i c u l a r distance o f Q (3 \hat{i} + 3 \hat{j} + 3 \hat{k}) f r o m t h e p l a n e \\ = | \frac{(3 \hat{i} + 3 \hat{j} + 3 \hat{k}) . (5 \hat{i} + 2 \hat{j} - 7 \hat{k}) + 9}{\sqrt[]{25 + 4 + 49}} | \\ = | \frac{15 + 6 - 21 + 9}{\sqrt[]{78}} | = | \frac{9}{\sqrt[]{78}} | \\ H e n c e, t h e t w o points a r e equidistant f r o m t h e g i v e n p l a n e . \\ O p p o s i t e s i g n s h o w s t h a t t h e y l i e o n e i t h e r s i d e o f t h e p l a n e . \end{array}$

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