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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

The angle between the planes $\vec{r} \cdot (2 \hat{i} - 3 \hat{j} + \hat{k}) = 1 and \vec{r} \cdot (\hat{i} - \hat{j}) = 4$ is $c o s^{- 1} (\frac{5}{\sqrt 58})$ .

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Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n p l a n e a r e \vec{r} . (2 \hat{i} - 3 \hat{j} + \hat{k}) = 1 a n d \vec{r} . (\hat{i} - \hat{j}) = 4 \\ H e r e, \vec{b_{1}} = 2 \hat{i} - 3 \hat{j} + \hat{k} a n d \vec{n_{2}} = \hat{i} - \hat{j} \\ S o, c o s θ = \frac{\vec{b_{1}} \vec{n_{2}}}{| \vec{b_{1}} | | \vec{n_{2}} |} \\ \Rightarrow c o s θ = \frac{(2 \hat{i} - 3 \hat{j} + \hat{k}) . (\hat{i} - \hat{j})}{\sqrt[]{4 + 9 + 1} \sqrt[]{1 + 1}} = \frac{2 + 3}{\sqrt[]{14} \sqrt[]{2}} = \frac{5}{\sqrt[]{28}} \\ ∴ θ = c o s^{- 1} (\frac{5}{\sqrt[]{28}}) w h i c h i s f a l s e . \\ H e n c e, t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

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 $s i n^{- 1} (5 / 2 \sqrt 91)$ . .

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Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} E q u a t i o n o f l i n e i s \vec{r} = (5 \hat{i} - \hat{j} - 4 \hat{k}) + λ (2 \hat{i} - \hat{j} + \hat{k}) a n d t h e \\ e q u a t i o n o f p l a n e i s \vec{r} . (3 \hat{i} - 4 \hat{j} - \hat{k}) + 5 = 0 \\ H e r e, \vec{b_{1}} = 2 \hat{i} - \hat{j} + \hat{k} a n d \vec{n_{2}} = 3 \hat{i} - 4 \hat{j} - \hat{k} \\ ∴ s i n θ = \frac{\vec{b_{1}} \vec{n_{2}}}{| \vec{b_{1}} | | \vec{n_{2}} |} \\ \Rightarrow s i n θ = \frac{(2 \hat{i} - \hat{j} + \hat{k}) . (3 \hat{i} - 4 \hat{j} - \hat{k})}{\sqrt[]{4 + 4 + 1} \sqrt[]{9 + 16 + 1}} = \frac{6 + 4 - 1}{\sqrt[]{6} \sqrt[]{26}} = \frac{9}{\sqrt[]{6} \sqrt[]{26}} \\ \Rightarrow s i n θ = \frac{9}{2 \sqrt[]{39}} w h i c h i s f a l s e . \\ H e n c e, t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

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

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Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

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

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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

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

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Payal Gupta

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar

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

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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

The cartesian equation of the plane $\hat{\vec{r}} \cdot \hat{i} + \hat{j} + \hat{k} = 2$ is __________.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n e q u a t i o n \vec{r} . (\hat{i} + \hat{j} - \hat{k}) = 2 \\ \Rightarrow (x \hat{i} + y \hat{j} + z \hat{k}) . (\hat{i} + \hat{j} - \hat{k}) = 2 \\ \Rightarrow x + y - z = 2 \\ H e n c e, t h e C a r t e s i a n e q u a t i o n o f t h e p l a n e i s x + y - z = 2 \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

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

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Payal Gupta

Contributor-Level 10

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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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

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

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Payal Gupta

Contributor-Level 10

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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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

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

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Payal Gupta

Contributor-Level 10

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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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

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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Payal Gupta

Contributor-Level 10

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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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

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) $\frac{3}{7}$

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Payal Gupta

Contributor-Level 10

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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