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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The vector equation of the line through the points (3, 4, –7) and (1, –1, 6) is __________.

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n points a r e (3, 4, - 7) a n d (1, - 1, 6) \\ H e r e, \vec{a} = 3 \hat{i} + 4 \hat{j} - 7 \hat{k} a n d \vec{b} = \hat{i} - \hat{j} + 6 \hat{k} \\ E q u a t i o n o f t h e l i n e i s \vec{r} = \vec{a} + λ (\vec{b} - \vec{a}) \\ \Rightarrow \vec{r} = (3 \hat{i} + 4 \hat{j} - 7 \hat{k}) + λ [(\hat{i} - \hat{j} + 6 \hat{k}) - (3 \hat{i} + 4 \hat{j} - 7 \hat{k})] \\ \Rightarrow \vec{r} = (3 \hat{i} + 4 \hat{j} - 7 \hat{k}) + λ (- 2 \hat{i} + - 5 \hat{j} + 13 \hat{k}) \\ \Rightarrow (x \hat{i} + y \hat{j} + z \hat{k}) = (3 \hat{i} + 4 \hat{j} - 7 \hat{k}) + λ (- 2 \hat{i} + - 5 \hat{j} + 13 \hat{k}) \\ \Rightarrow (x - 3) \hat{i} + (y - 4) \hat{j} + (z + 7) \hat{k} = λ (- 2 \hat{i} + - 5 \hat{j} + 13 \hat{k}) \\ H e n c e, t h e v e c t o r e q u a t i o n o f t h e l i n e i s \\ (x - 3) \hat{i} + (y - 4) \hat{j} + (z + 7) \hat{k} = λ (- 2 \hat{i} + - 5 \hat{j} + 13 \hat{k}) \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n e q u a t i o n i s \\ \frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} \\ H e r e, \vec{a} = (5 \hat{i} - 4 \hat{j} + 6 \hat{k}) a n d \vec{b} = (3 \hat{i} + 7 \hat{j} + 2 \hat{k}) \\ E q u a t i o n o f t h e l i n e i s \vec{r} = \vec{a} + \vec{b} λ \\ H e n c e, t h e v e c t o r e q u a t i o n o f t h e g i v e n l i n e i s \\ \vec{r} = (5 \hat{i} - 4 \hat{j} + 6 \hat{k}) + λ (3 \hat{i} + 7 \hat{j} + 2 \hat{k}) \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The plane $2 x - 3 y + 6 z - 11 = 0$ makes an angle $s i n^{- 1} (α)$ with the x-axis. The value of $α$ is equal to:

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

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

(C) $\frac{2}{7}$ 

(D) $\frac{3}{7}$

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} D i r e c t i o n r a t i o s o f t h e n o r m a l t o t h e p l a n e 2 x - 3 y + 6 z - 11 = 0 a r e 2, - 3, 6 \\ D i r e c t i o n r a t i o s o f x - a x i s a r e 1, 0, 0 \\ ∴ a n g l e b e t w e e n p l a n e a n d t h e l i n e i s \\ ∴ s i n θ = \frac{2 (1) - 3 (0) + 6 (0)}{\sqrt[]{{(2)}^{2} + {(- 3)}^{2} + {(6)}^{2}} . \sqrt[]{{(1)}^{2} + {(0)}^{2} + {(0)}^{2}}} \\ \Rightarrow s i n θ = \frac{2}{\sqrt[]{4 + 9 + 36}} = \frac{2}{7} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The locus represented by $x y + y z = 0$ is:

(A) A pair of perpendicular lines

(B) A pair of parallel lines

(C) A pair of parallel planes

(D) A pair of perpendicular planes

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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t x y + y z = 0 \\ y . (x + z) = 0 \\ y = 0 o r x + z = 0 \\ H e r e, y = 0 i s o n e p l a n e a n d x + z = 0 i s a n o t h e r p l a n e . S o, i t i s a p a i r o f p e r p e n d i c u l a r p l a n e s . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Exemplar Solution 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

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted•a year ago

Exemplar Solution 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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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

Exemplar Solution 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) $\frac{\sqrt 2}{10}$

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

Exemplar Solution 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

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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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