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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Find the vector equation of the line which is parallel to the vector $3 \hat{i} - 2 \hat{j} + 6 \hat{k}$ and which passes through the point (1, –2, 3).

2 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

$\begin{array}{l} W e k n o w t h a t t h e e q u a t i o n o f l i n e i s \\ \vec{r} = \vec{a} + \vec{b} λ \\ H e r e, \vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k} a n d \vec{b} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k} \\ ∴ E q u a t i o n o f l i n e i s \vec{r} = (\hat{i} - 2 \hat{j} + 3 \hat{k}) + λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) \\ \Rightarrow (x \hat{i} + y \hat{j} + z \hat{k}) = (\hat{i} - 2 \hat{j} + 3 \hat{k}) + λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) \\ \Rightarrow (x \hat{i} + y \hat{j} + z \hat{k}) - (\hat{i} - 2 \hat{j} + 3 \hat{k}) = λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) \\ \Rightarrow (x - 1) \hat{i} + (y + 2) \hat{j} + (z - 3) \hat{k} = λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s (x - 1) \hat{i} + (y + 2) \hat{j} + (z - 3) \hat{k} = λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Find the position vector of a point A in space such that $\vec{O A}$ is inclined at 60º to OX and at 45° to OY and $| \vec{O A} |$ = 10 units.

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

$\begin{array}{l} L e t α = 6 0^{0}, β = 4 5^{0} a n d t h e a n g l e i n c l i n e d t o O Z a x i s b e γ . \\ W e k n o w t h a t c o s^{2} α + c o s^{2} β + c o s^{2} γ = 1 \\ \Rightarrow c o s^{2} 6 0^{0} + c o s^{2} 4 5^{0} + c o s^{2} γ = 1 \\ \Rightarrow {(\frac{1}{2})}^{2} + {(\frac{1}{\sqrt[]{2}})}^{2} + c o s^{2} γ = 1 \Rightarrow \frac{1}{4} + \frac{1}{2} + c o s^{2} γ = 1 \\ \Rightarrow \frac{3}{4} + c o s^{2} γ = 1 \Rightarrow c o s^{2} γ = 1 - \frac{3}{4} = \frac{1}{4} \\ ∴ c o s γ = \pm \frac{1}{2} \Rightarrow c o s γ = \frac{1}{2} (Rejecting c o s γ = - \frac{1}{2}, since γ < 9 0^{0}) \\ ∴ \vec{O A} = | \vec{O A} | (\frac{1}{2} \hat{i} + \frac{1}{\sqrt[]{2}} \hat{j} + \frac{1}{2} \hat{k}) = 10 (\frac{1}{2} \hat{i} + \frac{1}{\sqrt[]{2}} \hat{j} + \frac{1}{2} \hat{k}) \\ = 5 \hat{i} + 5 \sqrt[]{2} \hat{j} + 5 \hat{k} \\ H e n c e, t h e p o s i t i o n v e c t o r o f A i s (5 \hat{i} + 5 \sqrt[]{2} \hat{j} + 5 \hat{k}) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

A vector  has magnitude 14 and direction ratios 2, 3, –6. Find the direction cosines and components of r?, given that  makes an acute angle with the x-axis.

8 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} L e t \vec{a}, \vec{b} a n d \vec{c} b e t h r e e v e c t o r s s u c h t h a t \vec{a} = 2 k, \vec{b} = 3 k a n d \vec{c} = - 6 k \\ I f l, m a n d n a r e t h e d i r e c t i o n s c o s i n e s o f v e c t o r \vec{r}, t h e n \\ l = \frac{\vec{a}}{| \vec{r} |} = \frac{2 k}{14} = \frac{k}{7} \\ m = \frac{\vec{b}}{| \vec{r} |} = \frac{3 k}{14} a n d n = \frac{\vec{c}}{| \vec{r} |} = \frac{- 6 k}{14} = \frac{- 3 k}{7} \\ W e k n o w t h a t l^{2} + m^{2} + n^{2} = 1 \Rightarrow \frac{k^{2}}{49} + \frac{9 k^{2}}{196} + \frac{9 k^{2}}{49} = 1 \\ \Rightarrow \frac{4 k^{2} + 9 k^{2} + 36 k^{2}}{196} = 1 \Rightarrow 49 k^{2} = 196 \Rightarrow k^{2} = 4 \\ ∴ k = \pm 2 a n d l = \frac{k}{7} = \frac{2}{7} \\ m = \frac{3 k}{14} = \frac{3 \times 2}{14} = \frac{3}{7} a n d n = \frac{- 3 k}{7} = \frac{- 3 \times 2}{7} = \frac{- 6}{7} \\ ∴ \hat{r} = \pm (\frac{2}{7} \hat{i} + \frac{3}{7} \hat{j} - \frac{6}{7} \hat{k}) \\ W e k n o w t h a t \vec{r} = (\hat{r}) | \vec{r} | \\ = \pm (\frac{2}{7} \hat{i} + \frac{3}{7} \hat{j} - \frac{6}{7} \hat{k}) . 14 = \pm (4 \hat{i} + 6 \hat{j} - 12 \hat{k}) \\ H e n c e, t h e r e q u i r e d d i r e c t i o n c o s i n e s a r e \frac{2}{7}, \frac{3}{7}, \frac{- 6}{7} a n d t h e c o m p o n e n t s o f \vec{r} a r e 4 \hat{i}, 6 \hat{j} a n d - 12 \hat{k} . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

A vector $\vec{r}$ . is inclined at equal angles to the three axes. If the magnitude of $r$ . is 2 $\sqrt[]{3}$ units, find $\vec{r}$ .

5 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} Since, t h e v e c t o r \vec{r} m a k e s e q u a l a n g l e s w i t h t h e a x e s, t h e i r d i r e c t i o n c o s i n e s s h o u l d b e s a m e . \\ ∴ l = m = n \\ W e k n o w t h a t l^{2} + m^{2} + n^{2} = 1 \Rightarrow l^{2} + l^{2} + l^{2} = 1 \\ \Rightarrow 3 l^{2} = 1 \Rightarrow l^{2} = \frac{1}{3} \Rightarrow l = \pm \frac{1}{\sqrt[]{3}} \\ ∴ \hat{r} = \pm \frac{1}{\sqrt[]{3}} \hat{i} \pm \frac{1}{\sqrt[]{3}} \hat{j} \pm \frac{1}{\sqrt[]{3}} \hat{k} \Rightarrow \hat{r} = \pm \frac{1}{\sqrt[]{3}} (\hat{i} + \hat{j} + \hat{k}) \\ W e k n o w t h a t \vec{r} = (\hat{r}) | \vec{r} | \\ = \pm \frac{1}{\sqrt[]{3}} (\hat{i} + \hat{j} + \hat{k}) 2 \sqrt[]{3} = \pm 2 (\hat{i} + \hat{j} + \hat{k}) \\ H e n c e, t h e r e q u i r e d v a l u e o f \vec{r} i s \pm 2 (\hat{i} + \hat{j} + \hat{k}) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

A vector <math> <mrow> <mover accent="true"> <mi>r</mi> <mo>→</mo> </mover> </mrow> </math> . is inclined at equal angles to the three axes. If the magnitude of <math> <mrow> <mi>r</mi> </mrow> </math> . is 2 <math> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> </math> units, find <math> <mrow> <mover accent="true"> <mi>r</mi> <mo>→</mo> </mover> </mrow> </math> .

1 Views|Posted a year ago

Asked by Alok Kumar Singh

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Using vectors, find the value of k such that the points (k, –10, 3), (1, –1, 3), and (3, 5, 3) are collinear.

7 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} L e t t h e g i v e n points a r e A (k, - 10, 3), B (1, - 1, 3) a n d C (3, 5, 3) \\ \vec{A B} = (1 - k) \hat{i} + (- 1 + 10) \hat{j} + (3 - 3) \hat{k} = (1 - k) \hat{i} + 9 \hat{j} + 0 \hat{k} \\ ∴ | \vec{A B} | = \sqrt[]{{(1 - k)}^{2} + {(9)}^{2}} = \sqrt[]{{(1 - k)}^{2} + 81} \\ \vec{B C} = (3 - 1) \hat{i} + (5 + 1) \hat{j} + (3 - 3) \hat{k} = 2 \hat{i} + 6 \hat{j} + 0 \hat{k} \\ ∴ | \vec{B C} | = \sqrt[]{{(2)}^{2} + {(6)}^{2}} = \sqrt[]{4 + 36} = \sqrt[]{40} = 2 \sqrt[]{10} \\ \vec{A C} = (3 - k) \hat{i} + (5 + 10) \hat{j} + (3 - 3) \hat{k} = (3 - k) \hat{i} + 6 \hat{j} + 0 \hat{k} \\ ∴ | \vec{A C} | = \sqrt[]{{(3 - k)}^{2} + {(15)}^{2}} = \sqrt[]{{(3 - k)}^{2} + 225} \\ I f A, B, a n d C a r e c o l l i n e a r, t h e n \\ | \vec{A B} | + | \vec{B C} | = | \vec{A C} | \\ \sqrt[]{{(1 - k)}^{2} + 81} + 2 \sqrt[]{10} = \sqrt[]{{(3 - k)}^{2} + 225} \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ {[\sqrt[]{{(1 - k)}^{2} + 81} + \sqrt[]{40}]}^{2} = {[\sqrt[]{{(3 - k)}^{2} + 225}]}^{2} \\ \Rightarrow {(1 - k)}^{2} + 81 + 40 + 2 \sqrt[]{40} \sqrt[]{{(1 - k)}^{2} + 81} = {(3 - k)}^{2} + 225 \\ \Rightarrow 1 + k^{2} - 2 k + 121 + 2 \sqrt[]{40} \sqrt[]{1 + k^{2} - 2 k + 81} = 9 + k^{2} - 6 k + 225 \\ \Rightarrow 122 - 2 k + 2 \sqrt[]{40} \sqrt[]{k^{2} - 2 k + 82} = 234 - 6 k \\ D i v i d i n g b y 2, w e g e t \\ \Rightarrow 61 - k + \sqrt[]{40} \sqrt[]{k^{2} - 2 k + 82} = 117 - 3 k \\ \Rightarrow \sqrt[]{40} \sqrt[]{k^{2} - 2 k + 82} = 117 - 61 - 3 k + k \\ \Rightarrow \sqrt[]{40} \sqrt[]{k^{2} - 2 k + 82} = 56 - 2 k \\ \Rightarrow 2 \sqrt[]{10} \sqrt[]{k^{2} - 2 k + 82} = 56 - 2 k \\ \Rightarrow \sqrt[]{10} \sqrt[]{k^{2} - 2 k + 82} = 28 - k (D i v i d i n g b y 2) \\ S q u a r i n g b o t h s i d e s, w e g e t \\ \Rightarrow 10 (k^{2} - 2 k + 82) = 784 + k^{2} - 56 k \\ \Rightarrow 10 k^{2} - 20 k + 820 = 784 + k^{2} - 56 k \\ \Rightarrow 10 k^{2} - k^{2} - 20 k + 56 k + 820 - 784 = 0 \\ \Rightarrow 9 k^{2} + 36 k + 36 = 0 \Rightarrow k^{2} + 4 k + 4 = 0 \\ \Rightarrow {(k + 2)}^{2} = 0 \Rightarrow k + 2 = 0 \Rightarrow k = - 2 \\ H e n c e, t h e r e q u i r e d v a l u e i s k = - 2 . \end{array}$

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Exemplar Solution 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.

5 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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Exemplar Solution 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.

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 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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Exemplar Solution 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 $\vec{a}$ − $\vec{b}$ .

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 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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Exemplar Solution 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.

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 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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Find the vector equation of the line which is parallel to the vector $3 \hat{i} - 2 \hat{j} + 6 \hat{k}$ and which passes through the point (1, –2, 3).

Find the position vector of a point A in space such that $\vec{O A}$ is inclined at 60º to OX and at 45° to OY and $| \vec{O A} |$ = 10 units.

A vector  has magnitude 14 and direction ratios 2, 3, –6. Find the direction cosines and components of r?, given that  makes an acute angle with the x-axis.

A vector $\vec{r}$ . is inclined at equal angles to the three axes. If the magnitude of $r$ . is 2 $\sqrt[]{3}$ units, find $\vec{r}$ .

Using vectors, find the value of k such that the points (k, –10, 3), (1, –1, 3), and (3, 5, 3) are collinear.

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.

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

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

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