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

If a variable line in two adjacent positions has direction cosines $l, m, n$ and $l + δ l, m + δ m, n + δ n$ ,show that the small angle $δ θ$ between the two positions is given by $δ θ^{2} = δ l^{2} + δ m^{2} + δ n^{2} .$

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} T h e g i v e n t h a t l, m, n a n d l + δ l, m + δ m, n + δ n a r e t h e d i r e c t i o n c o s i n e s o f a variable \\ l i n e i n t w o p o s i t i o n s \\ ∴ l^{2} + m^{2} + n^{2} = 1 \dots (i) \\ a n d {(l + δ l)}^{2} + {(m + δ m)}^{2} + {(n + δ n)}^{2} = 1 \dots (i i) \\ \Rightarrow l^{2} + δ l^{2} + 2 l . δ l + m^{2} + δ m^{2} + 2 m . δ m + n^{2} + δ n^{2} + 2 n . δ n = 1 \\ \Rightarrow (l^{2} + m^{2} + n^{2}) + (δ l^{2} + δ m^{2} + δ n^{2}) + 2 (l . δ l + m . δ m + n . δ n) = 1 \\ \Rightarrow 1 + (δ l^{2} + δ m^{2} + δ n^{2}) + 2 (l . δ l + m . δ m + n . δ n) = 1 \\ \Rightarrow l . δ l + m . δ m + n . δ n = - \frac{1}{2} (δ l^{2} + δ m^{2} + δ n^{2}) \\ L e t \vec{a} a n d \vec{b} b e t h e u n i t v e c t o r s a l o n g a l i n e w i t h d^{'} c o s i n e s l, m, n a n d (l + δ l), (m + δ m), (n + δ n) \\ ∴ \vec{a} = l \hat{i} + m \hat{j} + n \hat{k} a n d \vec{b} = (l + δ l) \hat{i} + (m + δ m) \hat{j} + (n + δ n) \hat{k} \\ ∴ c o s δ θ = \frac{\vec{a} . \vec{b}}{| \vec{a} | . | \vec{b} |} \\ \Rightarrow c o s δ θ = \frac{(l \hat{i} + m \hat{j} + n \hat{k}) [(l + δ l) \hat{i} + (m + δ m) \hat{j} + (n + δ n) \hat{k}]}{1.1} [? | \vec{a} | = | \vec{b} | = 1] \\ \Rightarrow c o s δ θ = l (l + δ l) + m (m + δ m) + n (n + δ n) \\ \Rightarrow c o s δ θ = l^{2} + l . δ l + m^{2} + m . δ m + n^{2} + n . δ n \\ \Rightarrow c o s δ θ = (l^{2} + m^{2} + n^{2}) + (l . δ l + m . δ m + n . δ n) \\ \Rightarrow c o s δ θ = 1 - \frac{1}{2} (δ l^{2} + δ m^{2} + δ n^{2}) \\ \Rightarrow 1 - c o s δ θ = \frac{1}{2} (δ l^{2} + δ m^{2} + δ n^{2}) \\ \Rightarrow 2 s i n^{2} \frac{δ θ}{2} = \frac{1}{2} (δ l^{2} + δ m^{2} + δ n^{2}) \\ \Rightarrow 4 s i n^{2} \frac{δ θ}{2} = δ l^{2} + δ m^{2} + δ n^{2} \\ \Rightarrow 4 {(\frac{δ θ}{2})}^{2} = δ l^{2} + δ m^{2} + δ n^{2} [? \frac{δ θ}{2} i s v e r y s m a l l s o, s i n \frac{δ θ}{2} = \frac{δ θ}{2}] \\ \Rightarrow {(δ θ)}^{2} = δ l^{2} + δ m^{2} + δ n^{2} H e n c e p r o v e d . \end{array}$

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

The vectors $λ \hat{i} + \hat{j} + 2 \hat{k}, \hat{i} + λ \hat{j} - \hat{k}, a n d 2 \hat{i} + \hat{j} + λ \hat{k}$ are coplanar if:

(A) $λ = - 2$

(B) $λ = 0$

(C) $λ = 1$

(D) $λ = - 1$

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alok kumar singh

Contributor-Level 10

This is an Objective type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t \vec{a} = λ \hat{i} + \hat{j} + 2 \hat{k} \\ ∴ \vec{b} = \hat{i} + λ \hat{j} - \hat{k} \\ \vec{c} = 2 \hat{i} - \hat{j} + λ \hat{k} \\ I f \vec{a}, \vec{b} a n d \vec{c} a r e c o p l a n a r, t h e n \\ \vec{a} . (\vec{b} * \vec{c}) = 0 \\ ∴ | \begin{matrix} λ & 1 & 2 \\ 1 & λ & - 1 \\ 2 & - 1 & λ \end{matrix} | = 0 \\ \Rightarrow λ (λ^{2} - 1) - 1 (λ + 2) + 2 (- 1 - 2 λ) = 0 \\ \Rightarrow λ^{3} - λ - λ - 2 - 2 - 4 λ = 0 \\ \Rightarrow λ^{3} - 6 λ - 4 = 0 \\ \Rightarrow (λ + 2) (λ^{2} - 2 λ - 2) = 0 \\ \Rightarrow λ = - 2 o r λ^{2} - 2 λ - 2 = 0 \\ \Rightarrow λ = \frac{2 \pm \sqrt[]{4 + 8}}{2} \\ \Rightarrow λ = \frac{2 \pm 2 \sqrt[]{3}}{2} \\ ∴ λ = - 2 o r 1 \pm \sqrt[]{3} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

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

Find the angle between the lines whose direction cosines are given by the equations $l + m + n = 0, l^{2} + m^{2} - n^{2} = 0 .$

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} T h e g i v e n e q u a t i o n s a r e \\ l + m + n = 0 \dots (i) \\ l^{2} + m^{2} + n^{2} = 0 \dots (i i) \\ F r o m e q u a t i o n (i) n = - (l + m) \\ P u t t i n g t h e v a l u e o f n i n e q . (i i) w e g e t \\ l^{2} + m^{2} + [- {(l + m)}^{2}] = 0 \\ \Rightarrow l^{2} + m^{2} - l^{2} - m^{2} - 2 l m = 0 \\ \Rightarrow - 2 l m = 0 \\ \Rightarrow l m = 0 \Rightarrow (- m - n) m = 0 [? l = - m - n] \\ \Rightarrow (m + n) m = 0 \Rightarrow m = 0 o r m = - n \\ \Rightarrow l = 0 o r l = - n \\ ∴ D i r e c t i o n c o s i n e s o f t h e t w o l i n e s a r e \\ 0, - n, n a n d - n, 0, n \Rightarrow 0, - 1, 1 a n d - 1, 0, 1 \\ ∴ c o s θ = \frac{(0 \hat{i} - \hat{j} + \hat{k}) . (- \hat{i} + 0 \hat{j} + \hat{k})}{\sqrt[]{{(- 1)}^{2} + {(1)}^{2}} \sqrt[]{{(- 1)}^{2} + {(1)}^{2}}} = \frac{1}{\sqrt[]{2} . \sqrt[]{2}} = \frac{1}{2} \\ ∴ θ = \frac{π}{3} \\ H e n c e, t h e r e q u i r e d a n g l e i s \frac{π}{3} . \end{array}$

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New answer posted

a year ago

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

If $∣ \vec{a} ∣ = 10$ , $∣ \vec{b} ∣ = 2$ , and $\vec{a} \cdot \vec{b} = 12$ , then the value of $∣ \vec{a} * \vec{b} ∣$ is:

(A) 5

(B) 10

(C) 14

(D) 16

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alok kumar singh

Contributor-Level 10

This is an Objective type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \vec{a} = 10, \vec{b} = 2 a n d \vec{a} . \vec{b} = 12 \\ ∴ \vec{a} . \vec{b} = | \vec{a} | | \vec{b} | c o s θ \\ \Rightarrow 12 = 10 . 2 . c o s θ \\ \Rightarrow c o s θ = \frac{12}{20} = \frac{3}{5} \\ ∴ s i n θ = \sqrt[]{1 - c o s^{2} θ} \\ \Rightarrow s i n θ = \sqrt[]{1 - {(\frac{3}{5})}^{2}} \Rightarrow s i n θ = \sqrt[]{1 - \frac{9}{25}} \\ \Rightarrow s i n θ = \sqrt[]{\frac{16}{25}} \Rightarrow s i n θ = \frac{4}{5} \\ N o w | \vec{a} * \vec{b} | = | \vec{a} | | \vec{b} | s i n θ \\ = 10 . 2 . \frac{4}{5} = 16 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New question posted

a year ago

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

If $? \vec{a} ? = 10$ , $? \vec{b} ? = 2$ , and $\vec{a} \cdot \vec{b} = 12$ , then the value of $? \vec{a} \times \vec{b} ?$ is:

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New answer posted

a year ago

Class 12th 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.

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Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Class 12th 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) $2 {\vec{a}}^{2}$

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alok kumar singh

Contributor-Level 10

This is an Objective type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th 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).

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Payal Gupta

Contributor-Level 10

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

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

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New question posted

a year ago

Class 12th 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).

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New answer posted

a year ago

Class 12th 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

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

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alok kumar singh

Contributor-Level 10

This is an Objective type Questions as classified in NCERT Exemplar

Sol:

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

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