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6 months ago

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

If $\vec{a}$ and $\vec{b}$ . are the position vectors of A and B, respectively, find the position vector of a point C in BA produced such that BC = 1.5 BA.

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alok kumar singh

Contributor-Level 10

This is a Short Answer type Questions as classified in NCERT Exempla

$\begin{array}{l} G i v e n t h a t B C = 1.5 B A \\ \Rightarrow \frac{B C}{B A} = 1.5 = \frac{3}{2} \\ \Rightarrow \frac{\vec{c} - \vec{b}}{\vec{a} - \vec{b}} = \frac{3}{2} \\ \Rightarrow 2 \vec{c} - 2 \vec{b} = 3 \vec{a} - 3 \vec{b} \Rightarrow 2 \vec{c} = 3 \vec{a} - 3 \vec{b} + 2 \vec{b} \\ \Rightarrow 2 \vec{c} = 3 \vec{a} - \vec{b} \\ ∴ \vec{c} = \frac{3 \vec{a} - \vec{b}}{2} \\ H e n c e, t h e r e q u i r e d v e c t o r i s \vec{c} = \frac{3 \vec{a} - \vec{b}}{2} . \end{array}$

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6 months ago

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

Find a unit vector in the direction of $\vec{PQ}$ , where P and Q have coordinates (5, 0, 8) and (3, 3, 2), respectively.

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alok kumar singh

Contributor-Level 10

This is a Short Answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n c o o r d i n a t e s a r e P (5, 0, 8) a n d Q (3, 3, 2) \\ ∴ \vec{P Q} = (3 - 5) \hat{i} + (3 - 0) \hat{j} + (2 - 8) \hat{k} = - 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \\ ∴ U n i t v e c t o r i n t h e d i r e c t i o n o f \vec{P Q} = \frac{\vec{P Q}}{| \vec{P Q} |} \\ = \frac{- 2 \hat{i} + 3 \hat{j} - 6 \hat{k}}{\sqrt[]{{(- 2)}^{2} + {(3)}^{2} + {(- 6)}^{2}}} = \frac{- 2 \hat{i} + 3 \hat{j} - 6 \hat{k}}{\sqrt[]{4 + 9 + 36}} \\ = \frac{- 2 \hat{i} + 3 \hat{j} - 6 \hat{k}}{\sqrt[]{49}} = \frac{- 2 \hat{i} + 3 \hat{j} - 6 \hat{k}}{7} \\ = \frac{1}{7} (- 2 \hat{i} + 3 \hat{j} - 6 \hat{k}) \\ H e n c e, t h e r e q u i r e d u n i t v e c t o r i s \frac{1}{7} (- 2 \hat{i} + 3 \hat{j} - 6 \hat{k}) . \end{array}$

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New answer posted

6 months ago

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

If â = î + ĵ + 2k̂ and b̂ = î + 2ĵ − 2k̂, find the unit vector in the direction of
(i) 6 $\vec{b}$
(ii) 2 $\vec{a}$ − $\vec{b}$ .

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alok kumar singh

Contributor-Level 10

This is a Short Answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \vec{a} = \hat{i} + \hat{j} + 2 \hat{k} a n d \vec{b} = 2 \hat{i} + \hat{j} - 2 \hat{k} \\ (i) 6 \vec{b} = 6 (2 \hat{i} + \hat{j} - 2 \hat{k}) = 12 \hat{i} + 6 \hat{j} - 12 \hat{k} \\ ∴ U n i t v e c t o r i n t h e d i r e c t i o n o f 6 \vec{b} = \frac{6 \vec{b}}{| 6 \vec{b} |} \\ = \frac{12 \hat{i} + 6 \hat{j} - 12 \hat{k}}{\sqrt[]{{(12)}^{2} + {(6)}^{2} + {(- 12)}^{2}}} = \frac{12 \hat{i} + 6 \hat{j} - 12 \hat{k}}{\sqrt[]{144 + 36 + 144}} \\ = \frac{12 \hat{i} + 6 \hat{j} - 12 \hat{k}}{\sqrt[]{324}} = \frac{12 \hat{i} + 6 \hat{j} - 12 \hat{k}}{18} \\ = \frac{6}{18} (2 \hat{i} + \hat{j} - 2 \hat{k}) = \frac{1}{3} (2 \hat{i} + \hat{j} - 2 \hat{k}) \\ H e n c e, t h e r e q u i r e d u n i t v e c t o r i s \frac{1}{3} (2 \hat{i} + \hat{j} - 2 \hat{k}) . \\ (i i) 2 \vec{a} - \vec{b} = 2 (\hat{i} + \hat{j} + 2 \hat{k}) - (2 \hat{i} + \hat{j} - 2 \hat{k}) \\ = 2 \hat{i} + 2 \hat{j} + 4 \hat{k} - 2 \hat{i} - \hat{j} + 2 \hat{k} = \hat{j} + 6 \hat{k} \\ ∴ U n i t v e c t o r i n t h e d i r e c t i o n o f 2 \vec{a} - \vec{b} = \frac{2 \vec{a} - \vec{b}}{| 2 \vec{a} - \vec{b} |} \\ = \frac{\hat{j} + 6 \hat{k}}{\sqrt[]{{(1)}^{2} + {(6)}^{2}}} = \frac{\hat{j} + 6 \hat{k}}{\sqrt[]{1 + 36}} \\ = \frac{\hat{j} + 6 \hat{k}}{\sqrt[]{37}} = \frac{1}{\sqrt[]{37}} (\hat{j} + 6 \hat{k}) \\ H e n c e, t h e r e q u i r e d u n i t v e c t o r i s \frac{1}{\sqrt[]{37}} (\hat{j} + 6 \hat{k}) . \end{array}$

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New answer posted

6 months ago

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

Find the unit vector in the direction of the sum of vectors $\vec{a}$ = î − 2ĵ + k̂ and $\vec{b}$ = ĵ + 2k̂.

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alok kumar singh

Contributor-Level 10

This is a Short Answer 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} - \hat{j} + \hat{k} a n d \vec{b} = 2 \hat{j} + \hat{k} \\ \vec{a} + \vec{b} = (2 \hat{i} - \hat{j} + \hat{k}) + (2 \hat{j} + \hat{k}) = 2 \hat{i} + \hat{j} + 2 \hat{k} \\ ∴ U n i t v e c t o r i n t h e d i r e c t i o n o f \vec{a} + \vec{b} = \frac{\vec{a} + \vec{b}}{| \vec{a} + \vec{b} |} \\ = \frac{2 \hat{i} + \hat{j} + 2 \hat{k}}{\sqrt[]{{(2)}^{2} + {(1)}^{2} + {(2)}^{2}}} = \frac{2 \hat{i} + \hat{j} + 2 \hat{k}}{\sqrt[]{4 + 1 + 4}} \\ = \frac{2 \hat{i} + \hat{j} + 2 \hat{k}}{\sqrt[]{9}} = \frac{2 \hat{i} + \hat{j} + 2 \hat{k}}{3} = \frac{2}{3} \hat{i} + \frac{1}{3} \hat{j} + \frac{2}{3} \hat{k} \\ H e n c e, t h e r e q u i r e d u n i t v e c t o r i s \frac{2}{3} \hat{i} + \frac{1}{3} \hat{j} + \frac{2}{3} \hat{k} . \end{array}

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New answer posted

7 months ago

Class 12th Cruise Culinary Academy

Does Cruise Culinary Academy accept Class 12th marks?

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Nitesh Gulati

Contributor-Level 10

Yes, Cruise Culinary Academy accept Class 12th marks for admission. The basic eligibility for course admission is to pass Class 12. Cruise Culinary Academy offers UG and Diploma couses under two streams, i.e., Business & Management Studies and Hospitality & Travel.

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New answer posted

7 months ago

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

Kindly consider the following

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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:

$\begin{array}{l} L e t I = \int_{- π}^{π} s i n^{3} x . c o s^{2} x d x \\ L e t f (x) = s i n^{3} x . c o s^{2} x \\ f (- x) = s i n^{3} (- x) . c o s^{2} (- x) = - s i n^{3} x . c o s^{2} x = - f (x) \\ ∴ \int_{- π}^{π} s i n^{3} x . c o s^{2} x d x i s a n o d d f u n c t i o n . \\ ∴ I = 0 \end{array}$

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New answer posted

7 months ago

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

Kindly consider the following

$\int \frac{s i n x}{3 + 4 c o s^{2} x} d x =_______$

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alok kumar singh

Contributor-Level 10

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

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New answer posted

7 months ago

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

Kindly consider the following

If: $\int^{a} \frac{1}{1 + 4 x^{2}} d x = \frac{π}{8}$ ,then, $a =$ _______.

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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:

$\begin{array}{l} L e t I = \int_{0}^{a} \frac{1}{1 + 4 x^{2}} d x = \frac{π}{8} \\ \Rightarrow \frac{1}{4} \int_{0}^{a} \frac{1}{(\frac{1}{4} + x^{2})} d x = \frac{π}{8} \Rightarrow \int_{0}^{a} \frac{1}{[{(\frac{1}{2})}^{2} + x^{2}]} d x = \frac{π}{2} \\ \Rightarrow \frac{1}{1 / 2} {[t a n^{- 1} \frac{x}{1 / 2}]}_{0}^{a} = \frac{π}{2} \Rightarrow 2 [t a n^{- 1} 2 a - t a n^{- 1} 0] = \frac{π}{2} \\ \Rightarrow t a n^{- 1} 2 a = \frac{π}{4} \Rightarrow 2 a = t a n \frac{π}{4} \Rightarrow 2 a = 1 \Rightarrow a = \frac{1}{2} \\ H e n c e, t h e v a l u e o f a = \frac{1}{2} . \end{array}$

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New answer posted

7 months ago

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

Kindly consider the following

$\int \frac{x + 3}{{(x + 4)}^{2}} e^{x} d x$ =_______.

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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:

$\begin{array}{l} L e t I = \int \frac{x + 3}{{(x + 4)}^{2}} . e^{x} d x \\ = \int \frac{x + 4 - 1}{{(x + 4)}^{2}} . e^{x} d x \\ = \int [\frac{x + 4}{{(x + 4)}^{2}} - \frac{1}{{(x + 4)}^{2}}] . e^{x} d x = \int [\frac{1}{x + 4} - \frac{1}{{(x + 4)}^{2}}] . e^{x} d x \\ P u t \frac{1}{x + 4} = t \Rightarrow - \frac{1}{{(x + 4)}^{2}} d x = d t \\ L e t f (x) = \frac{1}{x + 4} ∴ f^{'} (x) = - \frac{1}{{(x + 4)}^{2}} \\ Using \int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C \\ ∴ I = e^{x} . \frac{1}{x + 4} + C \\ H e n c e, I = \frac{e^{x}}{x + 4} + C . \end{array}$

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New answer posted

7 months ago

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

Kindly consider the following

$\int^{\frac{π}{2}} c o s x e^{s i n x} d x$ is equal to _______.

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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:

$\begin{array}{l} L e t I = \int_{0}^{\frac{π}{2}} c o s x . e^{s i n x} d x \\ P u t s i n x = t \Rightarrow c o s x d x = d t \\ W h e n x = 0 t h e n t = s i n 0 = 0 W h e n x = \frac{π}{2} t h e n t = s i n \frac{π}{2} = 1 \\ ∴ I = \int_{0}^{1} e^{t} d t = {[e^{t}]}_{0}^{1} = (e^{1} - e^{0}) = e - 1 \\ H e n c e, I = e - 1 \end{array}$

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