Page 99 - Exemplar Solution - Updates, Questions & Discussions on Exemplar Solution at Shiksha

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Find the equations of the line passing through the point (3, 0, 1) and parallel to the planes  $x + 2 y = 0$ and $3 y - z = 0$ .

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n point i s (3, 0, 1) a n d t h e e q u a t i o n o f p l a n e s a r e \\ x + 2 y = 0 \dots (i) \\ a n d 3 y - z = 0 \dots (i i) \\ E q u a t i o n o f a n y l i n e l passing t h r o u g h (3, 0, 1) i s \\ l : \frac{x - 3}{a} = \frac{y - 0}{b} = \frac{z - 1}{c} \\ D i r e c t i o n r a t i o s o f t h e n o r m a l t o t h e p l a n e (i) a n d (i i) a r e (1, 2, 0) a n d (0, 3, - 1) \\ Since, t h e l i n e i s p a r a l l e l t o b o t h t h e p l a n e s . \\ ∴ 1 . a + 2 . b + 0 . c = 0 \Rightarrow a + 2 b + 0 c = 0 \\ a n d 0 . a + 3 . b - 1 . c = 0 \Rightarrow 0 . a + 3 b - c = 0 \\ S o, \frac{a}{- 2 - 0} = \frac{- b}{- 1 - 0} = \frac{c}{3 - 0} = λ \\ ∴ a = - 2 λ, b = λ, c = 3 λ \\ S o, e q u a t i o n o f l i n e i s \\ \frac{x - 3}{- 2 λ} = \frac{y - 0}{λ} = \frac{z - 1}{3 λ} \\ H e n c e, t h e r e q u i r e d e q u a t i o n \frac{x - 3}{- 2} + \frac{y - 0}{1} + \frac{z - 1}{3} \\ o r i n v e c t o r f o r m i s (x - 3) \hat{i} + y \hat{j} + (z - 1) \hat{k} = λ (- 2 \hat{i} + \hat{j} + 3 \hat{k}) \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

The vectors $\vec{a}$ = 3î − 2ĵ + 2k̂ and b̂ = î − 2k̂ are the adjacent sides of a parallelogram. The acute angle between its diagonals is ________.

3 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n t h a t \vec{a} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k} \\ a n d \vec{b} = - \hat{i} - 2 \hat{k} \\ ∴ \vec{a} + \vec{b} = 2 \hat{i} - 2 \hat{j} a n d \vec{a} - \vec{b} = 4 \hat{i} - 2 \hat{j} + 4 \hat{k} \\ L e t θ b e t h e a n g l e b e t w e e n t h e t w o d i a g o n a l v e c t o r s \vec{a} + \vec{b} a n d \vec{a} - \vec{b} t h e n \\ c o s θ = \frac{(\vec{a} + \vec{b}) . (\vec{a} - \vec{b})}{| \vec{a} + \vec{b} | | \vec{a} - \vec{b} |} = \frac{(2 \hat{i} - 2 \hat{j}) . (4 \hat{i} - 2 \hat{j} + 4 \hat{k})}{\sqrt[]{{(2)}^{2} + {(- 2)}^{2}} \sqrt[]{{(4)}^{2} + {(- 2)}^{2} + {(4)}^{2}}} \\ = \frac{8 + 4}{2 \sqrt[]{2} . 6} = \frac{12}{2 \sqrt[]{2} . 6} = \frac{1}{\sqrt[]{2}} \\ ∴ θ = \frac{π}{4} \\ H e n c e, t h e v a l u e o f r e q u i r e d f i l l e r i s \frac{π}{4} . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n p l a n e i s 2 x - 2 y + 4 z + 5 = 0 a n d t h e g i v e n points i s (1, \frac{3}{2}, 2) \\ D' r a t i o s o f t h e n o r m a l t o t h e p l a n e a r e 2, - 2, 4 \\ S o, t h e ? e q u a t i o n o f l i n e passing t h r o u g h (1, \frac{3}{2}, 2) a n d w h o s e d' r a t i o s a r e e q u a l t o t h e \\ D' r a t i o s o f t h e n o r m a l t o t h e p l a n e i . e ., 2, - 2, 4 i s \frac{x - 1}{2} = \frac{y - \frac{3}{2}}{- 2} = \frac{z - 2}{4} = λ \\ ∴ A n y point i n t h e p l a n e i s 2 λ + 1, - 2 λ + \frac{3}{2}, 4 λ + 2 \\ Since, t h e point l i e s i n t h e p l a n e, t h e n \\ 2 (2 λ + 1) - 2 (- 2 λ + \frac{3}{2}) + 4 (4 λ + 2) + 5 = 0 \\ \Rightarrow 4 λ + 2 + 4 λ - 3 + 16 λ + 8 + 5 = 0 \\ \Rightarrow 24 λ + 12 = 0 ∴ λ = - \frac{1}{2} \\ S o, t h e c o o r d i n a t e s o f t h e point i n t h e p l a n e a r e \\ 2 (- \frac{1}{2}) + 1, - 2 (- \frac{1}{2}) + \frac{3}{2}, 4 (- \frac{1}{2}) + 2 i . e ., 0, \frac{5}{2}, 0 \\ H e n c e, t h e f o o t o f t h e p e r p e n d i c u l a r i s (0, \frac{5}{2}, 0) a n d t h e \\ r e q u i r e d l e n g t h = \sqrt[]{{(1 - 0)}^{2} + {(\frac{3}{2} - \frac{5}{2})}^{2} + {(2 - 0)}^{2}} \\ = \sqrt[]{1 + 1 + 4} = \sqrt[]{6} u n i t s \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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

18 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Find the distance of the point (2, 4, –1) from the line  <math> <mrow> <mfrac> <mrow> <mn>4</mn> <mo>−</mo> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>y</mi> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mo>−</mo> <mi>z</mi> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>.</mo> </mrow> </math>

5 Views|Posted a year ago

Asked by Payal Gupta

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If $\vec{r}$ . $\vec{a}$ = 0, $\vec{r}$ . $\vec{b}$ = 0, $\vec{r}$ . $\vec{c}$ = 0 for some non-zero vector r̂, then the value of
( $\vec{a}$ × $\vec{b}$ ) ĉ is ________.

3 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} I f \vec{r} i s a n o n - z e r o v e c t o r, t h e n \vec{a}, \vec{b} a n d \vec{c} c a n b e i n t h e s a m e p l a n e . \\ Since a n g l e s b e t w e e n \vec{a}, \vec{b} a n d \vec{c} a r e z e r o i . e . θ = 0 \\ ∴ \vec{a} . (\vec{b} \times \vec{c}) = 0 \\ H e n c e, t h e r e q u i r e d v a l u e i s 0 . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

The vector $\vec{a}$ + $\vec{b}$ bisects the angle between the non-collinear vectors $\vec{a}$ and $\vec{b}$ if ________.

3 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} I f v e c t o r \vec{a} + \vec{b} bisects t h e a n g l e b e t w e e n n o n - c o l l i n e a r v e c t o r s \vec{a} a n d \vec{b} t h e n t h e a n g l e b e t w e e n \\ \vec{a} + \vec{b} a n d \vec{a} i s e q u a l t o t h e a n g l e b e t w e e n \vec{a} + \vec{b} a n d \vec{b} . \\ S o, c o s θ = \frac{\vec{a} . (\vec{a} + \vec{b})}{| \vec{a} | | \vec{a} + \vec{b} |} = \frac{\vec{a} . (\vec{a} + \vec{b})}{| \vec{a} | \sqrt[]{a^{2} + b^{2}}} \dots (i) \\ A l s o, c o s θ = \frac{\vec{b} . (\vec{a} + \vec{b})}{| \vec{b} | | \vec{a} + \vec{b} |} [? θ i s s a m e] \\ = \frac{\vec{b} . (\vec{a} + \vec{b})}{| \vec{b} | \sqrt[]{a^{2} + b^{2}}} \dots (i i) \\ F r o m e q . (i) a n d e q . (i i) w e g e t, \\ \frac{\vec{a} . (\vec{a} + \vec{b})}{| \vec{a} | \sqrt[]{a^{2} + b^{2}}} = \frac{\vec{b} . (\vec{a} + \vec{b})}{| \vec{b} | \sqrt[]{a^{2} + b^{2}}} \\ \Rightarrow \frac{\vec{a}}{| \vec{a} |} = \frac{\vec{b}}{| \vec{b} |} \\ \Rightarrow \hat{a} = \hat{b} \Rightarrow \vec{a} = \vec{b} \\ H e n c e, t h e r e q u i r e d f i l l e r i s \vec{a} = \vec{b} . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

The number of unit vectors perpendicular to both $\vec{a} = 2 \hat{i} + 2 \hat{j} + \hat{k}, \vec{b} = \hat{j} + 2 \hat{k}$ is:

(A) One

(B) Two

(C) Three

(D) Infinite

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e n u m b e r o f v e c t o r s o f u n i t l e n g t h p e r p e n d i c u l a r t o v e c t o r s \vec{a} a n d \vec{b} i s \vec{c} (L e t) \\ ∴ \vec{c} = \pm (\vec{a} \times \vec{b}) \\ S o, t h e r e w i l l b e t w o v e c t o r s o f u n i t l e n g t h p e r p e n d i c u l a r t o v e c t o r s \vec{a} a n d \vec{b} . \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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.

8 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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.

4 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

Upvote

Exemplar Solution

Questions

Discussions

Active Users

Followers

Find the equations of the line passing through the point (3, 0, 1) and parallel to the planes  $x + 2 y = 0$ and $3 y - z = 0$ .

The vectors $\vec{a}$ = 3î − 2ĵ + 2k̂ and b̂ = î − 2k̂ are the adjacent sides of a parallelogram. The acute angle between its diagonals is ________.

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

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

If $\vec{r}$ . $\vec{a}$ = 0, $\vec{r}$ . $\vec{b}$ = 0, $\vec{r}$ . $\vec{c}$ = 0 for some non-zero vector r̂, then the value of
( $\vec{a}$ × $\vec{b}$ ) ĉ is ________.

The vector $\vec{a}$ + $\vec{b}$ bisects the angle between the non-collinear vectors $\vec{a}$ and $\vec{b}$ if ________.

The number of unit vectors perpendicular to both $\vec{a} = 2 \hat{i} + 2 \hat{j} + \hat{k}, \vec{b} = \hat{j} + 2 \hat{k}$ is:

(A) One

(B) Two

(C) Three

(D) Infinite

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 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.