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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If $∣ \vec{a} ∣ = 4$ and $- 3 \leq λ \leq 2$ , then the range of $∣ λ \vec{a} ∣$ is:

(A) [0, 8]

(B) [-12, 8]

(C) [0, 12]

(D) [8, 12]

13 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:

$\begin{array}{l} G i v e n t h a t | \vec{a} | = 4, - 3 \leq λ \leq 2 \\ N o w | λ \vec{a} | = λ | \vec{a} | = λ . 4 = 4 λ \\ H e r e, - 3 \leq λ \leq 2 \\ \Rightarrow - 3.4 \leq 4 λ \leq 2.4 \Rightarrow - 12 \leq 4 λ \leq 8 \\ ∴ 4 λ = [- 12, 8] \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If $\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $\vec{a} + \vec{b} + \vec{c} = 0$ and $? \vec{a} | = 2$ , $? \vec{b} | = 2, ? \vec{c} | = 2$ then the value of $\vec{a .} \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}$ is:

(A) 0

(B) 1

(C) $- 19$ 

(D) 38

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:

$\begin{array}{l} G i v e n t h a t | \vec{a} | = 2, | \vec{b} | = 3, | \vec{c} | = 5 \\ a n d \vec{a} + \vec{b} + \vec{c} = \vec{0} \\ ∴ (\vec{a} + \vec{b} + \vec{c}) . (\vec{a} + \vec{b} + \vec{c}) = \vec{0} . \vec{0} = 0 \\ {| \vec{a} |}^{2} + \vec{a} . \vec{b} + \vec{a} . \vec{c} + \vec{b} . \vec{a} + {| \vec{b} |}^{2} + \vec{b} . \vec{c} + \vec{c} . \vec{a} + \vec{c} . \vec{b} + {| \vec{c} |}^{2} = 0 \\ \Rightarrow {| \vec{a} |}^{2} + {| \vec{b} |}^{2} + {| \vec{c} |}^{2} + 2 \vec{a} . \vec{b} + 2 \vec{b} . \vec{c} + 2 \vec{c} . \vec{a} = 0 \\ \Rightarrow {(2)}^{2} + {(3)}^{2} + {(5)}^{2} + 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = 0 \\ \Rightarrow 4 + 9 + 25 + 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = 0 \\ \Rightarrow 38 + 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = 0 \\ \Rightarrow 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = - 38 \\ \Rightarrow \vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a} = - 19 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Two systems of rectangular axes have the same origin. If a plane cuts them at distances a, b, c and a′, b′, c′, respectively, from the origin, prove that $\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{a ’^{2}} + \frac{1}{b ’^{2}} + \frac{1}{c ’^{2}} .$

3 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

$\begin{array}{l} L e t O X, O Y, O Z a n d o x, o y, o z b e t w o rectangular s y s t e m s \\ ∴ E q u a t i o n s o f t w o p l a n e s a r e \\ \frac{X}{a}, \frac{Y}{b}, \frac{Z}{c} = 1 \dots (i) a n d \frac{x}{a^{'}}, \frac{y}{b^{'}}, \frac{z}{c^{'}} = 1 \dots (i i) \\ L e n g t h o f p e r p e n d i c u l a r f r o m o r i g i n t o p l a n e (i) i s \\ = \frac{\frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1}{\sqrt[]{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}} = \frac{1}{\sqrt[]{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}} \\ L e n g t h o f p e r p e n d i c u l a r f r o m o r i g i n t o p l a n e (i i) i s \\ = \frac{\frac{0}{a^{'}} + \frac{0}{b^{'}} + \frac{0}{c^{'}} - 1}{\sqrt[]{\frac{1}{{a^{'}}^{2}} + \frac{1}{{b^{'}}^{2}} + \frac{1}{{c^{'}}^{2}}}} = \frac{1}{\sqrt[]{\frac{1}{{a^{'}}^{2}} + \frac{1}{{b^{'}}^{2}} + \frac{1}{{c^{'}}^{2}}}} \\ A s p e r t h e c o n d i t i o n o f t h e q u e s t i o n \\ \frac{1}{\sqrt[]{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}}} = \frac{1}{\sqrt[]{\frac{1}{{a^{'}}^{2}} + \frac{1}{{b^{'}}^{2}} + \frac{1}{{c^{'}}^{2}}}} \\ H e n c e, \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{{a^{'}}^{2}} + \frac{1}{{b^{'}}^{2}} + \frac{1}{{c^{'}}^{2}} \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

The projection vector of $\vec{a .}$ on $\vec{b}$ is:

(A) $\frac{(\vec{a .} \cdot \vec{b})}{∣ \vec{b} ∣^{2}} \vec{b}$

(B) $\frac{\vec{a .} \cdot \vec{b}}{∣ \vec{b} ∣} \vec{b}$

(C) $\frac{(\vec{a .} \cdot \vec{b})}{∣ \vec{a .} ∣}$

(D) $\frac{(\vec{a} \cdot \vec{b})}{∣ \vec{a} ∣^{2}} \vec{b}$

6 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:

$\begin{array}{l} T h e p r o j e c t i o n v e c t o r o f \vec{a} o n \vec{b} = (\frac{\vec{a} . \vec{b}}{| \vec{b} |}) . \vec{b} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

O is the origin and A is (a, b, c). Find the direction cosines of the line OA and the equation of the plane through A at right angles to OA.

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

$\begin{array}{l} W e h a v e A (a, b, c) a n d O (0, 0, 0) \\ ∴ d i r e c t i o n r a t i o s o f O A = a - 0, b - 0, c - 0 \\ ∴ d i r e c t i o n c o s i n e s o f l i n e O A = \frac{a}{\sqrt[]{a^{2} + b^{2} + c^{2}}}, \frac{b}{\sqrt[]{a^{2} + b^{2} + c^{2}}}, \frac{c}{\sqrt[]{a^{2} + b^{2} + c^{2}}} \\ N o w, d i r e c t i o n r a t i o s o f t h e n o r m a l t o t h e p l a n e a r e (a, b, c) . \\ ∴ E q u a t i o n o f t h e p l a n e passing t h r o u g h t h e point A (a, b, c) i s \\ a (x - a) + b (y - b) + c (z - c) = 0 \\ \Rightarrow a x - a^{2} + b y - b^{2} + c z - c^{2} = 0 \\ \Rightarrow a x + b y + c z = a^{2} + b^{2} + c^{2} \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s a x + b y + c z = a^{2} + b^{2} + c^{2} . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = 0$ then the value of $\vec{a .} \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}$ is:

(A) 1

(B) 3

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

(D) None of these

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:

$\begin{array}{l} G i v e n t h a t | \vec{a} | = | \vec{b} | = | \vec{c} | = 1 \\ a n d \vec{a} + \vec{b} + \vec{c} = \vec{0} \\ ∴ (\vec{a} + \vec{b} + \vec{c}) . (\vec{a} + \vec{b} + \vec{c}) = \vec{0} . \vec{0} = 0 \\ {| \vec{a} |}^{2} + \vec{a} . \vec{b} + \vec{a} . \vec{c} + \vec{b} . \vec{a} + {| \vec{b} |}^{2} + \vec{b} . \vec{c} + \vec{c} . \vec{a} + \vec{c} . \vec{b} + {| \vec{c} |}^{2} = 0 \\ \Rightarrow {| \vec{a} |}^{2} + {| \vec{b} |}^{2} + {| \vec{c} |}^{2} + 2 \vec{a} . \vec{b} + 2 \vec{b} . \vec{c} + 2 \vec{c} . \vec{a} = 0 \\ \Rightarrow 1 + 1 + 1 + 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = 0 \\ \Rightarrow 2 (\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}) = - 3 \\ \Rightarrow \vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a} = \frac{- 3}{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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 $δ θ^{2} = δ l^{2} + δ m^{2} + δ n^{2} .$

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

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

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Exemplar Solution 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) $λ = - 1$

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:

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

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

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

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

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Exemplar Solution 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:

(A) 5

(B) 10

(C) 14

(D) 16

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:

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

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If $∣ \vec{a} ∣ = 4$ and $- 3 \leq λ \leq 2$ , then the range of $∣ λ \vec{a} ∣$ is:

(A) [0, 8]

(B) [-12, 8]

(C) [0, 12]

(D) [8, 12]

O is the origin and A is (a, b, c). Find the direction cosines of the line OA and the equation of the plane through A at right angles to OA.

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

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

(A) 5

(B) 10

(C) 14

(D) 16