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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Find the equation of a plane which bisects perpendicularly the line joining the points A(2, 3, 4) and B(4, 5, 8) at right angles.

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} T h e g i v e n t h a t A (2, 3, 4) a n d B (4, 5, 8) \\ C o o r d i n a t e s o f m i d - point C a r e (\frac{2 + 4}{2}, \frac{3 + 5}{2}, \frac{4 + 8}{2}) = (3, 4, 6) \\ N o w d i r e c t i o n r a t i o s o f t h e n o r m a l t o t h e p l a n e = d i r e c t i o n r a t i o s o f A B \\ = 4 - 2, 5 - 3, 8 - 4 = (2, 2, 4) \\ E q u a t i o n o f t h e p l a n e i s \\ a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0 \\ \Rightarrow 2 (x - 3) + 2 (y - 4) + 4 (z - 6) = 0 \\ \Rightarrow 2 x - 6 + 2 y - 8 + 4 z - 24 = 0 \\ \Rightarrow 2 x + 2 y + 4 z = 38 \\ \Rightarrow x + y + 2 z = 19 \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f p l a n e i s \\ x + y + 2 z = 19 o r \vec{r} (\hat{i} + \hat{j} + 2 \hat{k}) = 19 . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

The angle between two vectors $\vec{a}$ , $\vec{b}$ with magnitudes √3 and 4, respectively, and $\vec{a}$ . $\vec{b}$ = 2√3 is:

(A) $\frac{π}{6}$ 

(B) $\frac{π}{3}$ 

(C) $\frac{π}{2}$ 

(D) $\frac{5 π}{2}$

5 Views|Posted a year ago

Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, g i v e n t h a t | \vec{a} | = \sqrt[]{3}, | \vec{b} | = 4 a n d \vec{a} . \vec{b} = 2 \sqrt[]{3} \\ ∴ F r o m s c a l a r p r o d u c t, w e k n o w t h a t \\ \vec{a} . \vec{b} = | \vec{a} | | \vec{b} | c o s θ \\ \Rightarrow 2 \sqrt[]{3} = \sqrt[]{3} . 4 . c o s θ \\ \Rightarrow c o s θ = \frac{2 \sqrt[]{3}}{\sqrt[]{3} . 4} = \frac{1}{2} \\ ∴ θ = \frac{π}{3} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

The vector having initial and terminal points as (2, 5, 0) and (–3, 7, 4), respectively, is

(A) –î + 12ĵ + 4k̂

(B) 5î − 2ĵ − 4k̂

(C) −5î + 2ĵ + 4k̂

(D) î + ĵ + k̂

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 Objective type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t A a n d B b e t w o points w h o s e c o o r d i n a t e s a r e g i v e n a s (2, 5, 0) a n d (- 3, 7, 4) \\ ∴ \vec{A B} = (- 3 - 2) \hat{i} + (7 - 5) \hat{j} + (4 - 0) \hat{k} \\ \Rightarrow \vec{A B} = - 5 \hat{i} + 2 \hat{j} + 4 \hat{k} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Prove that the lines  $x = p y + q, z = r y + s$ and  $x = p ’ y + q ’, z = r ’ y + s ’$ are perpendicular if $p p ’ + r r ’ + 1 = 0$

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t : x = p y + q \Rightarrow y = \frac{x - q}{p} \\ a n d z = r y + s \Rightarrow y = \frac{z - s}{r} \\ ∴ t h e e q u a t i o n b e c o m e s \\ \frac{x - q}{p} = \frac{y}{1} = \frac{z - s}{r} i n w h i c h d^{'} r a t i o s a r e a_{1} = p, b_{1} = 1, c_{1} = r \\ S i m i l a r l y x = p^{'} y + q^{'} \Rightarrow y = \frac{x - q^{'}}{p^{'}} \\ a n d z = r^{'} y + s^{'} \Rightarrow y = \frac{z - s^{'}}{r^{'}} \\ ∴ t h e e q u a t i o n b e c o m e s \\ \frac{x - q^{'}}{p^{'}} = \frac{y}{1} = \frac{z - s^{'}}{r^{'}} i n w h i c h a_{2} = p^{'}, b_{2} = 1, c_{2} = r^{'} \\ I f t h e l i n e s a r e p e r p e n d i c u l a r t o e a c h o t h e r, t h e n \\ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 \\ p p^{'} + 1.1 + r r^{'} = 0 \\ H e n c e, t h e r e q u i r e d c o n d i t i o n i s p p^{'} + 1.1 + r r^{'} = 0 \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

The position vector of the point that divides the join of points 2â − 3b̂ and â + b̂ in the ratio 3:1 is:

(A) $\frac{(3 \vec{a} - 2 \hat{\vec{b}})}{2}$

(B) $\frac{(7 \vec{a} - 8 \hat{\vec{b}})}{4}$

(C) ^{$\frac{3 \vec{a}}{4}$}

(D) $\frac{5 \vec{a}}{4}$

5 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:

$\begin{array}{l} T h e g i v e n v e c t o r s a r e 2 \vec{a} - 3 \vec{b} a n d \vec{a} + \vec{b} a n d t h e r a t i o i s 3 : 1 . \\ ∴ T h e p o s i t i o n v e c t o r o f t h e r e q u i r e d point c w h i c h d i v i d e s t h e j o i n o f t h e g i v e n \\ v e c t o r s \vec{a} a n d \vec{b} i s \\ \vec{c} = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} = \frac{1 . (2 \vec{a} - 3 \vec{b}) + 3 (\vec{a} + \vec{b})}{3 + 1} = \frac{2 \vec{a} - 3 \vec{b} + 3 \vec{a} + 3 \vec{b}}{4} \\ = \frac{5 \vec{a}}{4} = \frac{5}{4} \vec{a} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Prove that the line through A(0, –1, –1) and B(4, 5, 1) intersects the line through C(3, 9, 4) and D(–4, 4, 4).

9 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} T h e g i v e n points a r e A (0, - 1, - 1) a n d B (4, 5, 1) \\ C (3, 9, 4) a n d D (- 4, 4, 4) \\ C a r t e s i a n f o r m o f e q u a t i o n A B i s \\ \frac{x - 0}{4 - 0} = \frac{y + 1}{5 + 1} = \frac{z + 1}{1 + 1} \Rightarrow \frac{x}{4} = \frac{y + 1}{6} = \frac{z + 1}{2} \\ a n d i t s v e c t o r f o r m i s \vec{r} = (- \hat{j} - \hat{k}) + λ (4 \hat{i} + 6 \hat{j} + 2 \hat{k}) \\ S i m i l a r l y, e q u a t i o n o f C D i s \\ \frac{x - 3}{- 4 - 3} = \frac{y - 9}{4 - 9} = \frac{z - 4}{4 - 4} \Rightarrow \frac{x - 3}{- 7} = \frac{y - 9}{- 5} = \frac{z - 4}{0} \\ a n d i t s v e c t o r f o r m i s \vec{r} = (3 \hat{i} + 9 \hat{j} + 4 \hat{k}) + μ (- 7 \hat{i} - 5 \hat{j}) \\ N o w, h e r e \vec{a_{1}} = - \hat{j} - \hat{k}, \vec{b_{1}} = 4 \hat{i} + 6 \hat{j} + 2 \hat{k} \\ a n d \vec{a_{2}} = 3 \hat{i} + 9 \hat{j} + 4 \hat{k}, \vec{b_{2}} = - 7 \hat{i} - 5 \hat{j} \\ S h o r t e s t distance b e t w e e n A B a n d C D \\ S . D . = | \frac{(\vec{a_{2}} - \vec{a_{1}}) (\vec{b_{1}} \times \vec{b_{2}})}{| \vec{b_{1}} \times \vec{b_{2}} |} | \\ \vec{a_{2}} - \vec{a_{1}} = (3 \hat{i} + 9 \hat{j} + 4 \hat{k}) - (- \hat{j} - \hat{k}) = 3 \hat{i} + 10 \hat{j} + 5 \hat{k} \\ \vec{b_{1}} \times \vec{b_{2}} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 6 & 2 \\ - 7 & - 5 & 0 \end{matrix} | \\ = \hat{i} (0 + 10) - \hat{j} (0 + 14) + \hat{k} (- 20 + 42) \\ = 10 \hat{i} - 14 \hat{j} + 22 \hat{k} \\ | \vec{b_{1}} \times \vec{b_{2}} | = \sqrt[]{{(10)}^{2} + {(- 14)}^{2} + {(22)}^{2}} = \sqrt[]{100 + 196 + 484} = \sqrt[]{780} \\ ∴ S . D = \frac{(3 \hat{i} + 10 \hat{j} + 5 \hat{k}) . (10 \hat{i} - 14 \hat{j} + 22 \hat{k})}{\sqrt[]{780}} = \frac{30 - 140 + 110}{\sqrt[]{780}} = 0 \\ H e n c e, t h e t w o l i n e s intersect e a c h o t h e r . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Choose the correct answer from the given four options in each of the Exercises 19 to 33:

The vector in the direction of the vector î − 2? + 2k? that has magnitude 9 is:

(A) î − 2ĵ + 2k̂

(B) $\frac{(\hat{i} - 2 \hat{j} + 2 \hat{k})}{3}$

(C) 9(î − 2ĵ + 2k̂)

(D) (î − 2ĵ + 2k̂)/3

4 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 Objective type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t \vec{a} = \hat{i} - 2 \hat{j} + 2 \hat{k} \\ U n i t v e c t o r i n t h e d i r e c t i o n o f \vec{a} = \frac{\vec{a}}{| \vec{a} |} \\ = \frac{\hat{i} - 2 \hat{j} + 2 \hat{k}}{\sqrt[]{{(1)}^{2} + {(- 2)}^{2} + {(2)}^{2}}} = \frac{\hat{i} - 2 \hat{j} + 2 \hat{k}}{\sqrt[]{1 + 4 + 4}} = \frac{\hat{i} - 2 \hat{j} + 2 \hat{k}}{3} \\ ∴ v e c t o r o f m a g n i t u d e 9 = \frac{9 (\hat{i} - 2 \hat{j} + 2 \hat{k})}{3} = 3 (\hat{i} - 2 \hat{j} + 2 \hat{k}) \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Find the angle between the lines $\vec{r} = (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) + λ (2 \hat{i} + \hat{j} + 6 \hat{k})$ and $\vec{r} = (2 \hat{i} + 5 \hat{k}) + μ (6 \hat{i} + 3 \hat{j} + 2 \hat{k}) .$

7 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

$\begin{array}{l} H e r e, \vec{b_{1}} = 2 \hat{i} + \hat{j} + 2 \hat{k} a n d \vec{b_{2}} = 6 \hat{i} + 3 \hat{j} + 2 \hat{k} \\ ∴ c o s θ = \frac{\vec{b_{1}} \vec{b_{2}}}{| \vec{b_{1}} | | \vec{b_{2}} |} = \frac{(2 \hat{i} + \hat{j} + 2 \hat{k}) . (6 \hat{i} + 3 \hat{j} + 2 \hat{k})}{\sqrt[]{{(2)}^{2} + {(1)}^{2} + {(2)}^{2}} . \sqrt[]{{(6)}^{2} + {(3)}^{2} + {(2)}^{2}}} \\ = \frac{12 + 3 + 4}{\sqrt[]{4 + 1 + 4} \sqrt[]{36 + 9 + 4}} = \frac{19}{\sqrt[]{9} \sqrt[]{49}} = \frac{19}{3.7} = \frac{19}{21} \\ ∴ θ = c o s^{- 1} (\frac{19}{21}) \\ H e n c e, t h e r e q u i r e d a n g l e i s c o s^{- 1} (\frac{19}{21}) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Show that the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = z$ intersect. Also, find their point of intersection.

3 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} T h e g i v e n e q u a t i o n s a r e \\ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} a n d \frac{x - 4}{5} = \frac{y - 1}{2} = z \\ L e t \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = λ \\ ∴ x = 2 λ + 1, y = 3 λ + 2 a n d z = 4 λ + 3 \\ a n d \frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1} = μ \\ ∴ x = 5 μ + 4, y = 2 μ + 1 a n d z = μ \\ I f t h e t w o l i n e s intersect e a c h o t h e r a t o n e point, \\ t h e n 2 λ + 1 = 5 μ + 4 \Rightarrow 2 λ - 5 μ = 3 \dots (i) \\ 3 λ + 2 = 2 μ + 1 \Rightarrow 3 λ - 2 μ = - 1 \dots (i i) \\ a n d 4 λ + 3 = μ \Rightarrow 4 λ - μ = - 3 \dots (i i i) \\ S o l v i n g e q n . (i) a n d (i i) w e g e t \\ 2 λ - 5 μ = 3 (M u l t i p l y b y 3) \\ 3 λ - 2 μ = - 1 (M u l t i p l y b y 2) \\ \Rightarrow 6 λ - 15 μ = 9 \\ 6 λ - 4 μ = - 2 \\ \underline{(-) (+) (+)} \\ - 11 μ = 11 ∴ μ = - 1 \\ P u t t i n g t h e v a l u e o f μ i n e q n . (i) w e g e t, \\ 2 λ - 5 (- 1) = 3 \\ \Rightarrow 2 λ + 5 = 3 \\ \Rightarrow 2 λ = - 2 ∴ λ = - 1 \\ N o w, p u t t i n g t h e v a l u e o f λ μ i n e q n . (i i i) t h e n, \\ 4 (- 1) - (- 1) = - 3 \\ - 4 + 1 = - 3 - 3 = - 3 (s a t i s f i e d) \\ ∴ C o o r d i n a t e s o f t h e point o f intersection a r e \\ x = 5 (- 1) + 4 = - 5 + 4 = - 1 \\ y = 2 (- 1) + 1 = - 2 + 1 = - 1 \\ z = - 1 \\ H e n c e, t h e g i v e n l i n e s intersect e a c h o t h e r a t (- 1, - 1, - 1) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If $\vec{a}$ = î + ĵ + k̂ and b̂ = ĵ − k̂, find a vector $c$ such that $\vec{a} \times \vec{c}$ = $\vec{b}$ and $\vec{a} . \vec{c}$ = 3.

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 Long Answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} \\ A l s o g i v e n t h a t \vec{a} = \hat{i} + \hat{j} + \hat{k} a n d \vec{b} = \hat{j} - \hat{k} \\ S i n c e, \vec{a} \times \vec{c} = \vec{b} \\ ∴ | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ c_{1} & c_{2} & c_{3} \end{matrix} | = \hat{j} - \hat{k} \\ \hat{i} (c_{3} - c_{2}) - \hat{j} (c_{3} - c_{1}) + \hat{k} (c_{2} - c_{1}) = \hat{j} - \hat{k} \\ O n c o m p a r i n g t h e l i k e t e r m s, w e g e t \\ c_{3} - c_{2} = 0 \dots (i) \\ c_{1} - c_{3} = 1 \dots (i i) \\ c_{2} - c_{1} = - 1 \dots (i i i) \\ N o w f o r \vec{a} . \vec{c} = 3 \\ (\hat{i} + \hat{j} + \hat{k}) . (c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}) = 3 \\ ∴ c_{1} + c_{2} + c_{3} = 3 \dots (i v) \\ A d d i n g e q . (i i) a n d e q . (i i i) w e g e t, \\ c_{2} - c_{3} = 0 \dots (v) \\ F r o m (i v) a n d (v) w e g e t \\ c_{1} + 2 c_{2} = 3 \dots (v i) \\ F r o m (i i i) a n d (v i) w e g e t \\ c_{1} + 2 c_{2} = 3 \\ \underline{- c_{1} + c_{2} = - 1} \\ A d d i n g 3 c_{2} = 2 \\ ∴ c_{2} = \frac{2}{3} \\ c_{3} - c_{2} = 0 \Rightarrow c_{3} - \frac{2}{3} = 0 ∴ c_{3} = \frac{2}{3} \\ N o w, c_{2} - c_{1} = - 1 \Rightarrow \frac{2}{3} - c_{1} = - 1 \\ \Rightarrow c_{1} = 1 + \frac{2}{3} = \frac{5}{3} \\ ∴ \vec{c} = \frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k} = \frac{1}{3} (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) \\ H e n c e, \vec{c} = \frac{1}{3} (5 \hat{i} + 2 \hat{j} + 2 \hat{k}) . \end{array}$

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Find the equation of a plane which bisects perpendicularly the line joining the points A(2, 3, 4) and B(4, 5, 8) at right angles.

The vector having initial and terminal points as (2, 5, 0) and (–3, 7, 4), respectively, is

(A) –î + 12ĵ + 4k̂

(B) 5î − 2ĵ − 4k̂

(C) −5î + 2ĵ + 4k̂

(D) î + ĵ + k̂

Prove that the lines  $x = p y + q, z = r y + s$ and  $x = p ’ y + q ’, z = r ’ y + s ’$ are perpendicular if $p p ’ + r r ’ + 1 = 0$

The position vector of the point that divides the join of points 2â − 3b̂ and â + b̂ in the ratio 3:1 is:

(A) $\frac{(3 \vec{a} - 2 \hat{\vec{b}})}{2}$

(B) $\frac{(7 \vec{a} - 8 \hat{\vec{b}})}{4}$

(C) ^{$\frac{3 \vec{a}}{4}$}

(D) $\frac{5 \vec{a}}{4}$

Prove that the line through A(0, –1, –1) and B(4, 5, 1) intersects the line through C(3, 9, 4) and D(–4, 4, 4).

Choose the correct answer from the given four options in each of the Exercises 19 to 33:

The vector in the direction of the vector î − 2? + 2k? that has magnitude 9 is:

(A) î − 2ĵ + 2k̂

(B) $\frac{(\hat{i} - 2 \hat{j} + 2 \hat{k})}{3}$

(C) 9(î − 2ĵ + 2k̂)

(D) (î − 2ĵ + 2k̂)/3

Find the angle between the lines $\vec{r} = (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) + λ (2 \hat{i} + \hat{j} + 6 \hat{k})$ and $\vec{r} = (2 \hat{i} + 5 \hat{k}) + μ (6 \hat{i} + 3 \hat{j} + 2 \hat{k}) .$

Show that the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = z$ intersect. Also, find their point of intersection.

If $\vec{a}$ = î + ĵ + k̂ and b̂ = ĵ − k̂, find a vector $c$ such that $\vec{a} \times \vec{c}$ = $\vec{b}$ and $\vec{a} . \vec{c}$ = 3.