Exemplar Solution

Get insights from 1.3k questions on Exemplar Solution, answered by students, alumni, and experts. You may also ask and answer any question you like about Exemplar Solution

1.3k
Questions
0
Discussions
0
Active Users
0
Followers

Have a question about Exemplar Solution? Ask our experts

Ask Your Question

Keep it short & simple. Type complete word. Avoid abusive language.

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Find the area of the region bounded by the curves $y^{2} = 9 x$ and $y = 3 x$ .

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} W e h a v e, y^{2} = 9 x, y = 3 x \\ S o l v i n g t h e t w o e q u a t i o n s, w e h a v e \\ {(3 x)}^{2} = 9 x \\ \Rightarrow 9 x^{2} - 9 x = 0 \Rightarrow 9 x (x - 1) = 0 \\ ∴ x = 0, 1 \\ A r e a o f t h e s h a d e d r e g i o n \\ = a r (r e g i o n O A B) - a r (Δ O A B) \\ = \int_{0}^{1} y_{1} d x = \int_{0}^{1} \sqrt[]{9 x} d x - \int_{0}^{1} 3 x d x \\ = 3 \int_{0}^{1} \sqrt[]{x} d x - 3 \int_{0}^{1} x d x = 3 \times \frac{2}{3} {[x^{3 / 2}]}_{0}^{1} - 3 {[\frac{x^{2}}{2}]}_{0}^{1} \\ = 2 [{(1)}^{3 / 2} - 0] - \frac{3}{2} [{(1)}^{2} - 0] = 2 (1) - \frac{3}{2} (1) = 2 - \frac{3}{2} = \frac{1}{2} s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = \frac{1}{2} s q . u n i t s . \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

If the foot of the perpendicular drawn from the origin to a plane is (5, –3, –2), then the equation of the plane is $\vec{r} \cdot (5 \hat{i} - 3 \hat{j} + 2 \hat{k}) - 38 .$

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} T h e g i v e n e q u a t i o n o f t h e p l a n e i s \vec{r} . (5 \hat{i} - 3 \hat{j} - 2 \hat{k}) = 38 \\ I f t h e f o o t o f t h e p e r p e n d i c u l a r t o t h i s p l a n e i s \\ (5, - 3, - 2) i . e ., 5 \hat{i} - 3 \hat{j} - 2 \hat{k} t h e n \\ \Rightarrow (5 \hat{i} - 3 \hat{j} - 2 \hat{k}) . (5 \hat{i} - 3 \hat{j} - 2 \hat{k}) = 38 \\ \Rightarrow 25 + 9 + 4 = 38 \\ 38 = 38 (s a t i s f i e d) \\ H e n c e, t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The equation of a line, which is parallel to $2 \hat{i} + \hat{j} + \hat{3 k}$ and passes through the point (5, –2, 4), is $\frac{x - 5}{2} = \frac{y + 2}{- 1} = \frac{z - 4}{3} .$

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} H e r e, x_{1} = 5, y_{1} = - 2, z_{1} = 4; a = 2, b = 1, c = 3 \\ W e k n o w t h a t t h e e q u a t i o n o f l i n e i s \frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c} \\ \Rightarrow \frac{x - 5}{2} = \frac{y + 2}{1} = \frac{z - 4}{3} \\ H e n c e, t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

Upvote

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 $\hat{r} = 5 \hat{i} - 4 \hat{j} + 6 \hat{k} + λ (3 \hat{i} + 7 \hat{j} + 2 \hat{k}) .$

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} T h e C a r t e s i a n f o r m o f t h e e q u a t i o n i s \\ \frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} = λ \\ H e r e, x_{1} = 5, y_{1} = - 4, z_{1} = 6; a = 3, b = 7, c = 2 \\ S o, t h e v e c t o r e q u a t i o n i s \vec{r} = (5 \hat{i} - 4 \hat{j} + 6 \hat{k}) + λ (3 \hat{i} + 7 \hat{j} + 2 \hat{k}) \\ H e n c e, t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The line $\vec{r} = \hat{2 i} + \hat{3 j} + \hat{k} + λ (\hat{i} - \hat{j} + \hat{k})$ lies in the plane $\hat{r} \cdot (3 \hat{i} + \hat{j} + \hat{k}) + 2 = 0 .$

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} D i r e c t i o n r a t i o s o f t h e l i n e (\hat{i} - \hat{j} + 2 \hat{k}) \\ D i r e c t i o n r a t i o s o f t h e n o r m a l t o t h e p l a n e a r e (3 \hat{i} + \hat{j} - \hat{k}) \\ S o, (\hat{i} - \hat{j} + 2 \hat{k}) . (3 \hat{i} + \hat{j} - \hat{k}) = 3 - 1 - 2 = 0 \\ T h e r e f o r e, t h e l i n e i s p a r a l l e l t o t h e p l a n e . \\ N o w point t h r o u g h w h i c h t h e l i n e i s passing \to =2 \hat{i} - 3 \hat{j} - \hat{k} \\ I f t h e l i n e l i e s i n t h e p l a n e t h e n \\ (2 \hat{i} - 3 \hat{j} - \hat{k}) . (3 \hat{i} + \hat{j} - \hat{k}) + 2 = 0 \\ 6 - 3 + 1 + 2 \neq 0 \\ S o, t h e l i n e d o e s n o t l i e i n t h e p l a n e . \\ H e n c e, t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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 $c o s^{- 1} (\frac{5}{\sqrt 58})$ .

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

Upvote

New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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

2 Views|Posted a year ago

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

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

Upvote

Exemplar Solution

Questions

Discussions

Active Users

Followers