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New answer posted

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

The vector $\vec{a}$ + $\vec{b}$ bisects the angle between the non-collinear vectors $\vec{a}$ and $\vec{b}$ if ________.

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alok kumar singh

Contributor-Level 10

This is a Fill in the blanks type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} I f v e c t o r \vec{a} + \vec{b} bisects t h e a n g l e b e t w e e n n o n - c o l l i n e a r v e c t o r s \vec{a} a n d \vec{b} t h e n t h e a n g l e b e t w e e n \\ \vec{a} + \vec{b} a n d \vec{a} i s e q u a l t o t h e a n g l e b e t w e e n \vec{a} + \vec{b} a n d \vec{b} . \\ S o, c o s θ = \frac{\vec{a} . (\vec{a} + \vec{b})}{| \vec{a} | | \vec{a} + \vec{b} |} = \frac{\vec{a} . (\vec{a} + \vec{b})}{| \vec{a} | \sqrt[]{a^{2} + b^{2}}} \dots (i) \\ A l s o, c o s θ = \frac{\vec{b} . (\vec{a} + \vec{b})}{| \vec{b} | | \vec{a} + \vec{b} |} [? θ i s s a m e] \\ = \frac{\vec{b} . (\vec{a} + \vec{b})}{| \vec{b} | \sqrt[]{a^{2} + b^{2}}} \dots (i i) \\ F r o m e q . (i) a n d e q . (i i) w e g e t, \\ \frac{\vec{a} . (\vec{a} + \vec{b})}{| \vec{a} | \sqrt[]{a^{2} + b^{2}}} = \frac{\vec{b} . (\vec{a} + \vec{b})}{| \vec{b} | \sqrt[]{a^{2} + b^{2}}} \\ \Rightarrow \frac{\vec{a}}{| \vec{a} |} = \frac{\vec{b}}{| \vec{b} |} \\ \Rightarrow \hat{a} = \hat{b} \Rightarrow \vec{a} = \vec{b} \\ H e n c e, t h e r e q u i r e d f i l l e r i s \vec{a} = \vec{b} . \end{array}$

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New answer posted

a year ago

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

The number of unit vectors perpendicular to both $\vec{a} = 2 \hat{i} + 2 \hat{j} + \hat{k}, \vec{b} = \hat{j} + 2 \hat{k}$ is:

(A) One

(B) Two

(C) Three

(D) Infinite

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alok kumar singh

Contributor-Level 10

This is an Objective type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e n u m b e r o f v e c t o r s o f u n i t l e n g t h p e r p e n d i c u l a r t o v e c t o r s \vec{a} a n d \vec{b} i s \vec{c} (L e t) \\ ∴ \vec{c} = \pm (\vec{a} * \vec{b}) \\ S o, t h e r e w i l l b e t w o v e c t o r s o f u n i t l e n g t h p e r p e n d i c u l a r t o v e c t o r s \vec{a} a n d \vec{b} . \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

a year ago

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

Find the foot of the perpendicular from the point (2, 3, –8) to the line $\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .$ Also, find the perpendicular distance from the given point to the line.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t \frac{4 - x}{2} + \frac{y}{6} + \frac{1 - z}{3} i s t h e e q u a t i o n o f l i n e \\ \Rightarrow \frac{x - 4}{- 2} + \frac{y}{6} + \frac{z - 1}{- 3} = λ \\ ∴ C o o r d i n a t e s o f a n y point Q o n t h e l i n e a r e \\ x = - 2 λ + 4, y = 6 λ a n d z = - 3 λ + 1 \\ a n d t h e g i v e n point i s P (2, 3, - 8) \\ D i r e c t i o n r a t i o s o f P Q a r e - 2 λ + 4 - 2, 6 λ - 3, - 3 λ + 1 + 8 \\ a n d t h e D' r a t i o s o f t h e g i v e n l i n e a r e - 2, 6, - 3 . \\ I f P Q ⊥ l i n e \\ t h e n - 2 (- 2 λ + 2) + 6 (6 λ - 3) - 3 (- 3 λ + 9) = 0 \\ \Rightarrow 4 λ - 4 + 36 λ - 18 + 9 λ - 27 = 0 \\ \Rightarrow 49 λ - 49 = 0 \Rightarrow λ = 1 \\ ∴ T h e f o o t o f t h e p e r p e n d i c u l a r i s - 2 (1) + 4, 6 (1), - 3 (1) + 1 \\ i . e ., 2, 6, - 2 \\ N o w, distance P Q = \sqrt[]{{(2 - 2)}^{2} + {(3 - 6)}^{2} + {(- 8 + 2)}^{2}} \\ = \sqrt[]{9 + 36} = \sqrt[]{45} = 3 \sqrt[]{5} \\ H e n c e, t h e r e q u i r e d c o o r d i n a t e s o f t h e f o o t o f t h e p e r p e n d i c u l a r a r e 2, 6, - 2 a n d t h e \\ r e q u i r e d distance 3 \sqrt[]{5} u n i t s . \end{array}$

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New answer posted

a year ago

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

Find the foot of the perpendicular from the point (2, 3, –8) to the line $\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .$ Also, find the perpendicular distance from the given point to the line.

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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

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New answer posted

a year ago

Class 12th 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]

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alok kumar singh

Contributor-Level 10

This is an Objective type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th 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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an Objective type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th 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 $\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{a'^{2}} + \frac{1}{b'^{2}} + \frac{1}{c'^{2}} .$

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Payal Gupta

Contributor-Level 10

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

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

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New answer posted

a year ago

Class 12th 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) $\frac{(\vec{a} \cdot \vec{b})}{∣ \vec{a} ∣^{2}} \vec{b}$

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alok kumar singh

Contributor-Level 10

This is an Objective type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th 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.

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Class 12th Exemplar Solution

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Objective type Questions as classified in NCERT Exemplar

Sol:

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

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