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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If <math> <mrow> <mo>?</mo> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> <mo>?</mo> <mo>=</mo> <mn>1</mn> <mn>0</mn> </mrow> </math> , <math> <mrow> <mo>?</mo> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> <mo>?</mo> <mo>=</mo> <mn>2</mn> </mrow> </math> , and <math> <mrow> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> <mo>⋅</mo> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> <mo>=</mo> <mn>1</mn> <mn>2</mn> </mrow> </math> , then the value of <math> <mrow> <mo>?</mo> <mover accent="true"> <mi>a</mi> <mo>→</mo> </mover> <mo>×</mo> <mover accent="true"> <mi>b</mi> <mo>→</mo> </mover> <mo>?</mo> </mrow> </math> is:

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Find the equations of the two lines through the origin which intersect the line $\frac{x - 3}{2} = \frac{y - 1}{1} = \frac{z}{1}$ at angles of $\frac{π}{3}$ each.

4 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

$\begin{array}{l} A n y point o n t h e g i v e n l i n e i s \\ \frac{x - 3}{2} = \frac{y - 3}{1} = \frac{z}{1} = λ \\ \Rightarrow x = 2 λ + 3, y = λ + 3 a n d z = λ \\ L e t i t b e t h e c o o r d i n a t e s o f P \\ ∴ D i r e c t i o n r a t i o s o f O P a r e (2 λ + 3 - 0), (λ + 3 - 0) a n d (λ - 0) \\ \Rightarrow 2 λ + 3, λ + 3, λ \\ B u t t h e d i r e c t i o n s r a t i o s o f t h e l i n e P Q a r e 2, 1, 1 \\ ∴ c o s θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt[]{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt[]{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}} \\ \Rightarrow c o s \frac{π}{3} = \frac{2 (2 λ + 3) + 1 (λ + 3) + 1 . λ}{\sqrt[]{{(2)}^{2} + {(1)}^{2} + {(1)}^{2}} . \sqrt[]{{(2 λ + 3)}^{2} + {(λ + 3)}^{2} + {(λ)}^{2}}} \\ \Rightarrow \frac{1}{2} = \frac{4 λ + 6 + λ + 3 + λ}{\sqrt[]{4 λ^{2} + 9 + 12 λ + λ^{2} + 9 + 6 λ + λ^{2}}} \\ \Rightarrow \frac{\sqrt[]{6}}{2} = \frac{6 λ + 9}{\sqrt[]{6 λ^{2} + 18 λ + 18}} = \frac{6 λ + 9}{\sqrt[]{6} \sqrt[]{λ^{2} + 3 λ + 3}} \\ \Rightarrow \frac{6}{2} = \frac{3 (2 λ + 3)}{\sqrt[]{λ^{2} + 3 λ + 3}} \Rightarrow 3 = \frac{3 (2 λ + 3)}{\sqrt[]{λ^{2} + 3 λ + 3}} \\ \Rightarrow 1 = \frac{2 λ + 3}{\sqrt[]{λ^{2} + 3 λ + 3}} \Rightarrow \sqrt[]{λ^{2} + 3 λ + 3} = 2 λ + 3 \\ \Rightarrow λ^{2} + 3 λ + 3 = 4 λ^{2} + 12 λ + 9 (S q u a r i n g b o t h s i d e s) \\ \Rightarrow 3 λ^{2} + 9 λ + 6 = 0 \Rightarrow λ^{2} + 3 λ + 2 = 0 \\ \Rightarrow (λ + 1) (λ + 2) = 0 ∴ λ = - 1, λ = - 2 \\ ∴ D i r e c t i o n s r a t i o s a r e [2 (- 1) + 3, - 1 + 3, - 1] i . e ., 1, 2, - 1 w h e n λ = - 1 a n d \\ [2 (- 2) + 3, - 2 + 3, - 2] i . e ., - 1, 1, - 2 w h e n λ = - 2 . \\ H e n c e, t h e r e q u i r e d e q u a t i o n s a r e \frac{x}{1} = \frac{y}{2} = \frac{z}{- 1} a n d \frac{x}{- 1} = \frac{y}{1} = \frac{z}{- 2} \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

For any vector , the value of  is equal to:

(A) ${\vec{a}}^{2}$ 

(B) $3 {\vec{a}}^{2}$ 

(C) $4 {\vec{a}}^{2}$ 

(D) $2 {\vec{a}}^{2}$

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:

$\begin{array}{l} L e t \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \\ ∴ {\vec{a}}^{2} = a_{1}^{2} + a_{2}^{2} + a_{3}^{2} \\ N o w, \vec{a} \times \hat{i} = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) \times \hat{i} \\ = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ 1 & 0 & 0 \end{matrix} | \\ = \hat{i} (0 - 0) - \hat{j} (0 - a_{3}) + \hat{k} (0 - a_{2}) = a_{3} \hat{j} - a_{2} \hat{k} \\ ∴ {(\vec{a} \times \hat{i})}^{2} = (a_{3} \hat{j} - a_{2} \hat{k}) . (a_{3} \hat{j} - a_{2} \hat{k}) = a_{3}^{2} + a_{2}^{2} \\ S i m i l a r l y {(\vec{a} \times \hat{j})}^{2} = a_{1}^{2} + a_{3}^{2} a n d {(\vec{a} \times \hat{k})}^{2} = a_{1}^{2} + a_{2}^{2} \\ ∴ {(\vec{a} \times \hat{i})}^{2} + {(\vec{a} \times \hat{j})}^{2} + {(\vec{a} \times \hat{k})}^{2} = a_{3}^{2} + a_{2}^{2} + a_{1}^{2} + a_{3}^{2} + a_{1}^{2} + a_{2}^{2} \\ = 2 (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) = 2 {\vec{a}}^{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Find the equation of the plane through the points (2, 1, 0), (3, –2, –2), and (3, 1, 7).

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

$\begin{array}{l} Since, t h e e q u a t i o n o f t h e p l a n e passing t h r o u g h t h e p o i n t s (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) a n d (x_{3}, y_{3}, z_{3}) i s \\ \Rightarrow | \begin{matrix} x - x_{1} & y - y_{1} & z - z_{1} \\ x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ x_{3} - x_{1} & y_{3} - y_{1} & z_{3} - z_{1} \end{matrix} | = 0 \\ \Rightarrow | \begin{matrix} x - 2 & y - 1 & z - 0 \\ 3 - 2 & - 2 - 1 & - 2 - 0 \\ 3 - 2 & 1 - 1 & 7 - 0 \end{matrix} | = 0 \\ \Rightarrow | \begin{matrix} x - 2 & y - 1 & z \\ 1 & - 3 & - 2 \\ 1 & 0 & 7 \end{matrix} | = 0 \\ \Rightarrow (x - 2) | \begin{matrix} - 3 & - 2 \\ 0 & 7 \end{matrix} | - (y - 1) | \begin{matrix} 1 & - 2 \\ 1 & 7 \end{matrix} | + z | \begin{matrix} 1 & - 3 \\ 1 & 0 \end{matrix} | = 0 \\ \Rightarrow (x - 2) (- 21) - (y - 1) (7 + 2) + z (3) = 0 \\ \Rightarrow - 21 (x - 2) - 9 (y - 1) + 3 z = 0 \\ \Rightarrow - 21 x + 42 - 9 y + 9 + 3 z = 0 \\ \Rightarrow - 21 x - 9 y + 3 z + 51 = 0 \Rightarrow 7 x + 3 y - z - 17 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s 7 x + 3 y - z - 17 = 0 \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Find the equation of the plane through the points (2, 1, 0), (3, –2, –2), and (3, 1, 7).

1 Views|Posted a year ago

Asked by Payal Gupta

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

The vectors from the origin to the points A and B are: respectively. Then the area of triangle OAB is:

(A) 340

(B) 25

(C) 229

(D) $\frac{1}{2} \sqrt[]{229}$

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:

$\begin{array}{l} L e t O b e t h e o r i g i n \\ ∴ \vec{O A} = 2 \hat{i} - 3 \hat{j} + 2 \hat{k} a n d \vec{O B} = 2 \hat{i} - 3 \hat{j} + 2 \hat{k} \\ ∴ A r e a o f Δ O A B = \frac{1}{2} | \vec{O A} \times \vec{O B} | = \frac{1}{2} | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 3 & 2 \\ 2 & 3 & 1 \end{matrix} | \\ = \frac{1}{2} | \hat{i} (- 3 - 6) - \hat{j} (2 - 4) + \hat{k} (6 + 6) | \\ = \frac{1}{2} | - 9 \hat{i} + 2 \hat{j} + 12 \hat{k} | = \frac{1}{2} \sqrt[]{{(- 9)}^{2} + {(2)}^{2} + {(12)}^{2}} \\ = \frac{1}{2} \sqrt[]{81 + 4 + 144} = \frac{1}{2} \sqrt[]{229} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter 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) $\frac{2}{5}$

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 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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Exemplar Solution 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) − $\frac{5}{2}$

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 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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Exemplar Solution 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.

2 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

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

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Exemplar Solution 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.

4 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

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

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Find the equations of the two lines through the origin which intersect the line $\frac{x - 3}{2} = \frac{y - 1}{1} = \frac{z}{1}$ at angles of $\frac{π}{3}$ each.

Find the equation of the plane through the points (2, 1, 0), (3, –2, –2), and (3, 1, 7).

<p><strong>Find the equation of the plane through the points (2, 1, 0), (3, –2, –2), and (3, 1, 7).</strong></p>

The vectors from the origin to the points A and B are: respectively. Then the area of triangle OAB is:

(A) 340

(B) 25

(C) 229

(D) $\frac{1}{2} \sqrt[]{229}$

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) − $\frac{5}{2}$

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.

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.