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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

The value of λ for which the vectors 3î − 6? + k? and 2î − 4? + λk? are parallel is:

(A) $\frac{2}{3}$

(B) $\frac{3}{2}$

(C) $\frac{5}{2}$

(D) $\frac{2}{5}$

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is an Objective type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} a n d \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} \\ I f \vec{a} ? \vec{b} \\ ∴ \frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} = \frac{a_{3}}{b_{3}} \\ \Rightarrow \frac{3}{2} = \frac{- 6}{- 4} = \frac{1}{λ} \Rightarrow \frac{1}{λ} = \frac{3}{2} \Rightarrow λ = \frac{2}{3} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Find the value of λ such that the vectors $\vec{a}$ = 2î + λĵ + k̂ and $\vec{b}$ = î + 2ĵ + 3k̂ are orthogonal.

(A) 0

(B) 1

(C) $\frac{3}{2}$

(D) − $\frac{5}{2}$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an Objective type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \vec{a} = 2 \hat{i} + λ \hat{j} + \hat{k} a n d \vec{b} = \hat{i} + 2 \hat{j} + 3 \hat{k} \\ Since \vec{a} a n d \vec{b} a r e o r t h o g o n a l \\ ∴ \vec{a} . \vec{b} = 0 \\ \Rightarrow (2 \hat{i} + λ \hat{j} + \hat{k}) . (\hat{i} + 2 \hat{j} + 3 \hat{k}) = 0 \\ \Rightarrow 2 + 2 λ + 3 = 0 \\ \Rightarrow 5 + 2 λ = 0 \Rightarrow λ = \frac{- 5}{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

If the line drawn from the point (–2, –1, –3) meets a plane at right angles at the point (1, –3, 3), find the equation of the plane.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} D i r e c t i o n r a t i o s o f t h e n o r m a l t o t h e p l a n e a r e \\ = (1 + 2, - 3 + 1, 3 + 3) = (3, - 2, 6) \\ E q u a t i o n o f t h e p l a n e passing t h r o u g h o n e point (x_{1}, y_{1}, z_{1}) i s \\ a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0 \\ \Rightarrow 3 (x - 1) - 2 (y + 3) + 6 (z - 3) = 0 \\ \Rightarrow 3 x - 3 - 2 y - 6 + 6 z - 18 = 0 \\ \Rightarrow 3 x - 2 y + 6 z = 27 \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f p l a n e i s 3 x - 2 y + 6 z = 27 \end{array}$

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Find the equation of a plane which is at a distance of $3 \sqrt[]{3}$ units from the origin and the normal to which is equally inclined to coordinate axes.

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} Since, t h e n o r m a l t o t h e p l a n e i s e q u a l l y i n c l i n e d t o t h e a x i s \\ ∴ c o s α = c o s β = c o s γ \\ \Rightarrow c o s^{2} α + c o s^{2} α + c o s^{2} α = 1 \\ \Rightarrow 3 c o s^{2} α = 1 \Rightarrow c o s α = \frac{1}{\sqrt[]{3}} \\ \Rightarrow c o s α = c o s β = c o s γ = \frac{1}{\sqrt[]{3}} \\ S o, t h e n o r m a l i s \vec{N} = \frac{1}{\sqrt[]{3}} \hat{i} + \frac{1}{\sqrt[]{3}} \hat{j} + \frac{1}{\sqrt[]{3}} \hat{k} \\ ∴ E q u a t i o n o f t h e p l a n e i s \vec{r} . \vec{N} = d \\ \Rightarrow \vec{r} . \frac{\vec{N}}{| \vec{N} |} = d \\ \Rightarrow \frac{\vec{r} . (\frac{1}{\sqrt[]{3}} \hat{i} + \frac{1}{\sqrt[]{3}} \hat{j} + \frac{1}{\sqrt[]{3}} \hat{k})}{1} = 3 \sqrt[]{3} \\ \Rightarrow \vec{r} . (\frac{1}{\sqrt[]{3}} \hat{i} + \frac{1}{\sqrt[]{3}} \hat{j} + \frac{1}{\sqrt[]{3}} \hat{k}) = 3 \sqrt[]{3} \\ \Rightarrow (x \hat{i} + y \hat{j} + z \hat{k}) . \frac{1}{\sqrt[]{3}} (\hat{i} + \hat{j} + \hat{k}) = 3 \sqrt[]{3} \\ \Rightarrow x + y + z = 3 \sqrt[]{3} . \sqrt[]{3} = 9 \\ H e n c e, t h e r e q u i r e d e q u a t i o n o f p l a n e i s x + y + z = 9 . \end{array}$

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New answer posted

a year ago

Class 12th 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.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Class 12th 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) $\frac{5 π}{2}$

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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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New answer posted

a year ago

Class 12th 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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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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New answer posted

a year ago

Class 12th 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 $p p' + r r' + 1 = 0$

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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

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

0 0

New answer posted

a year ago

Class 12th 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) $\frac{5 \vec{a}}{4}$

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

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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New answer posted

a year ago

Class 12th 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).

0 Follower 9 Views

Payal Gupta

Contributor-Level 10

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

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

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