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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

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.

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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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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

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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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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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

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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Payal Gupta

Contributor-Level 10

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

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

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Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

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).

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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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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

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 $\vec{r} = (2 \hat{i} + 5 \hat{k}) + μ (6 \hat{i} + 3 \hat{j} + 2 \hat{k}) .$

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Payal Gupta

Contributor-Level 10

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

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

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.

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Payal Gupta

Contributor-Level 10

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

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

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

Find the vector equation of the line which is parallel to the vector $3 \hat{i} - 2 \hat{j} + 6 \hat{k}$ and which passes through the point (1, –2, 3).

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} W e k n o w t h a t t h e e q u a t i o n o f l i n e i s \\ \vec{r} = \vec{a} + \vec{b} λ \\ H e r e, \vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k} a n d \vec{b} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k} \\ ∴ E q u a t i o n o f l i n e i s \vec{r} = (\hat{i} - 2 \hat{j} + 3 \hat{k}) + λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) \\ \Rightarrow (x \hat{i} + y \hat{j} + z \hat{k}) = (\hat{i} - 2 \hat{j} + 3 \hat{k}) + λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) \\ \Rightarrow (x \hat{i} + y \hat{j} + z \hat{k}) - (\hat{i} - 2 \hat{j} + 3 \hat{k}) = λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) \\ \Rightarrow (x - 1) \hat{i} + (y + 2) \hat{j} + (z - 3) \hat{k} = λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s (x - 1) \hat{i} + (y + 2) \hat{j} + (z - 3) \hat{k} = λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) . \end{array}$

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New answer posted

3 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Eleven Exemplar Solution

Find the position vector of a point A in space such that $\vec{O A}$ is inclined at 60º to OX and at 45° to OY and $| \vec{O A} |$ = 10 units.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} L e t α = 6 0^{0}, β = 4 5^{0} a n d t h e a n g l e i n c l i n e d t o O Z a x i s b e γ . \\ W e k n o w t h a t c o s^{2} α + c o s^{2} β + c o s^{2} γ = 1 \\ \Rightarrow c o s^{2} 6 0^{0} + c o s^{2} 4 5^{0} + c o s^{2} γ = 1 \\ \Rightarrow {(\frac{1}{2})}^{2} + {(\frac{1}{\sqrt[]{2}})}^{2} + c o s^{2} γ = 1 \Rightarrow \frac{1}{4} + \frac{1}{2} + c o s^{2} γ = 1 \\ \Rightarrow \frac{3}{4} + c o s^{2} γ = 1 \Rightarrow c o s^{2} γ = 1 - \frac{3}{4} = \frac{1}{4} \\ ∴ c o s γ = \pm \frac{1}{2} \Rightarrow c o s γ = \frac{1}{2} (Rejecting c o s γ = - \frac{1}{2}, since γ < 9 0^{0}) \\ ∴ \vec{O A} = | \vec{O A} | (\frac{1}{2} \hat{i} + \frac{1}{\sqrt[]{2}} \hat{j} + \frac{1}{2} \hat{k}) = 10 (\frac{1}{2} \hat{i} + \frac{1}{\sqrt[]{2}} \hat{j} + \frac{1}{2} \hat{k}) \\ = 5 \hat{i} + 5 \sqrt[]{2} \hat{j} + 5 \hat{k} \\ H e n c e, t h e p o s i t i o n v e c t o r o f A i s (5 \hat{i} + 5 \sqrt[]{2} \hat{j} + 5 \hat{k}) . \end{array}$

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