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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Show that the area of the parallelogram whose diagonals are given by $\vec{a}$ and $\vec{b}$ is $\frac{1}{2} ∣ \frac{\vec{a} \times \vec{b}}{2} ∣$ Also, find the area of the parallelogram whose diagonals are 2î − ĵ + k̂ and î + 3ĵ − k̂.

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

\begin{array}{l} L e t A B C D b e a parallelogram s u c h t h a t, \\ \vec{A B} = \vec{p}, \vec{A D} = \vec{q} = \vec{B C} \\ ∴ b y l a w o f t r i a n g l e, w e g e t \\ \vec{A C} = \vec{a} = \vec{p} + \vec{q} \dots (i) \\ a n d \vec{B D} = \vec{b} = - \vec{p} + \vec{q} \dots (i i) \\ A d d i n g e q . (i) a n d (i i) w e g e t, \\ \vec{a} + \vec{b} = 2 \vec{q} \Rightarrow \vec{q} = (\frac{\vec{a} + \vec{b}}{2}) \\ S u b t r a c t i n g e q . (i i) f r o m (i) w e g e t, \\ \vec{a} - \vec{b} = 2 \vec{p} \Rightarrow \vec{p} = (\frac{\vec{a} - \vec{b}}{2}) \\ ∴ \vec{p} \times \vec{q} = \frac{1}{4} (\vec{a} + \vec{b}) (\vec{a} - \vec{b}) = \frac{1}{4} (\vec{a} \times \vec{a} - \vec{a} \times \vec{b} + \vec{b} \times \vec{a} - \vec{b} \times \vec{b}) \\ = \frac{1}{4} (- \vec{a} \times \vec{b} + \vec{b} \times \vec{a}) [\begin{array}{l} ? \vec{a} \times \vec{a} = 0 \\ \vec{b} \times \vec{b} = 0 \end{array}] \\ = \frac{1}{4} (\vec{a} \times \vec{b} + \vec{b} \times \vec{a}) = \frac{1}{4} . 2 (\vec{a} \times \vec{b}) = \frac{| \vec{a} \times \vec{b} |}{2} \\ S o, t h e a r e a o f parallelogram A B C D = | \vec{p} \times \vec{q} | = \frac{1}{2} | \vec{a} \times \vec{b} | \\ N o w a r e a o f parallelogram w h o s e d i a g o n a l s a r e 2 \hat{i} - \hat{j} + \hat{k} a n d \hat{i} + 3 \hat{j} - \hat{k} \\ = \frac{1}{2} | (2 \hat{i} - \hat{j} + \hat{k}) \times (\hat{i} + 3 \hat{j} - \hat{k}) | \\ = \frac{1}{2} | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 1 & 1 \\ 1 & 3 & - 1 \end{matrix} | \\ = \frac{1}{2} [\hat{i} (1 - 3) - \hat{j} (- 2 - 1) + \hat{k} (6 + 1)] = \frac{1}{2} | - 2 \hat{i} + 3 \hat{j} + 7 \hat{k} | \\ = \frac{1}{2} \sqrt[]{{(- 2)}^{2} + {(3)}^{2} + {(7)}^{2}} = \frac{1}{2} \sqrt[]{4 + 9 + 49} = \frac{1}{2} \sqrt[]{62} s q . u n i t s . \\ H e n c e, t h e r e q u i r e d a r e a i s \frac{1}{2} \sqrt[]{62} s q . u n i t s . \end{array}

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If $\vec{a}, \vec{b}, \vec{c}$ determine the vertices of a triangle, show that $\frac{1}{2} (\vec{b} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{b})$ gives the $\vec{a}, \vec{b}, \vec{c}$ vector area of the triangle. Hence deduce the condition that the three points are collinear. Also, find the unit vector normal to the plane of the triangle.

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

$\begin{array}{l} Since, \vec{a}, \vec{b} a n d \vec{c} a r e t h e v e r t i c e s o f Δ A B C \\ ∴ \vec{A B} = \vec{b} - \vec{a}, \vec{B C} = \vec{c} - \vec{b} a n d \vec{A c} = \vec{c} - \vec{a} \\ ∴ A r e a o f Δ A B C = \frac{1}{2} | \vec{A B} \times \vec{A C} | \\ = \frac{1}{2} | (\vec{b} - \vec{a}) \times (\vec{c} - \vec{b}) | = \frac{1}{2} | \vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{a} \times \vec{c} + \vec{a} \times \vec{a} | \\ = \frac{1}{2} | \vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a} | [\begin{array}{l} ? \vec{a} \times \vec{b} = - \vec{b} \times \vec{a} \\ \vec{c} \times \vec{a} = - \vec{a} \times \vec{c} \\ \vec{a} \times \vec{a} = \vec{0} \end{array}] \\ F o r t h r e e v e c t o r s a r e c o l l i n e a r, a r e a o f Δ A B C = 0 \\ ∴ \frac{1}{2} | \vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a} | = 0 \\ | \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} | = 0 \\ w h i c h i s t h e c o n d i t i o n o f c o l l i n e a r i t y o f \vec{a}, \vec{b} a n d \vec{c} . \\ L e t \hat{n} b e t h e u n i t v e c t o r n o r m a l t o t h e p l a n e o f t h e Δ A B C \\ ∴ \hat{n} = \frac{\vec{A B} \times \vec{A C}}{| \vec{A B} \times \vec{A C} |} = \frac{\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}}{| \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} |} \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Prove that in any triangle ABC, $c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$ where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively.

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Asked by Alok Kumar Singh

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} H e r e, i n t h e g i v e n f i g u r e, t h e c o m p o n e n t s o f c a r e c c o s A a n d a n d c s i n A . \\ ∴ \vec{C D} = b - c c o s A \\ I n Δ B D C, \\ a^{2} = C D^{2} + B D^{2} \\ \Rightarrow a^{2} = {(b - c c o s A)}^{2} + {(c s i n A)}^{2} \\ \Rightarrow a^{2} = b^{2} + c^{2} c o s^{2} A - 2 b c c o s A + c^{2} s i n^{2} A \\ \Rightarrow a^{2} = b^{2} + c^{2} (c o s^{2} A + s i n^{2} A) - 2 b c c o s A \\ \Rightarrow a^{2} = b^{2} + c^{2} - 2 b c c o s A \\ \Rightarrow 2 b c c o s A = b^{2} + c^{2} - a^{2} \\ ∴ c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \\ H e n c e p r o v e d . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Using vectors, prove that parallelograms on the same base and between the same parallels are equal in area.

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t A B C D a n d A B F E b e t w o parallelograms o n t h e s a m e b a s e A B a n d b e t w e e n s a m e \\ p a r a l l e l l i n e s A B a n d D F . \\ L e t \vec{A B} = \vec{a} a n d \vec{A D} = \vec{b} \\ ∴ A r e a o f parallelogram A B C D = | \vec{a} \times \vec{b} | \\ N o w A r e a o f parallelogram A B F E = | \vec{A B} \times \vec{A E} | \\ = | \vec{a} \times (\vec{A D} + \vec{D E}) | = | \vec{a} \times (\vec{b} + K \vec{a}) | \\ = | (\vec{a} \times \vec{b}) + K (\vec{a} \times \vec{a}) | = | \vec{a} \times \vec{b} | + 0 [? \vec{a} \times \vec{a} = 0] \\ = | \vec{a} \times \vec{b} | \\ H e n c e p r o v e d . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, –1, 4), C(4, 5, –1).

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n t h a t A (1, 2, 3), B (2, - 1, 4) a n d C (4, 5, - 1) \\ \vec{A B} = (2 - 1) \hat{i} + (- 1 - 2) \hat{j} + (4 - 3) \hat{k} \\ \vec{A B} = \hat{i} - 3 \hat{j} + \hat{k} \\ \vec{A C} = (4 - 1) \hat{i} + (5 - 2) \hat{j} + (- 1 - 3) \hat{k} = 3 \hat{i} + 3 \hat{j} - 4 \hat{k} \\ A r e a o f Δ A B C = \frac{1}{2} | \vec{A B} \times \vec{A C} | = \frac{1}{2} | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 3 & 1 \\ 3 & 3 & - 4 \end{matrix} | \\ = \frac{1}{2} [\hat{i} (12 - 3) - \hat{j} (- 4 - 3) + \hat{k} (3 + 9)] = \frac{1}{2} | 9 \hat{i} + 7 \hat{j} + 12 \hat{k} | \\ = \frac{1}{2} \sqrt[]{{(9)}^{2} + {(7)}^{2} + {(12)}^{2}} = \frac{1}{2} \sqrt[]{81 + 49 + 144} = \frac{1}{2} \sqrt[]{274} \\ H e n c e, t h e r e q u i r e d a r e a i s \frac{\sqrt[]{274}}{2} . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If A, B, C, D are the points with position vectors î + ĵ − k̂, 2î − 3ĵ + k̂, 2î − 3k̂, and 3î + 2ĵ − k̂, respectively, find the projection of $\bar{A B}$ along $\bar{C D .}$

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

$\begin{array}{l} H e r e, P o s i t i o n v e c t o r o f A = \hat{i} + \hat{j} - \hat{k} \\ P o s i t i o n v e c t o r o f B = 2 \hat{i} - \hat{j} + 3 \hat{k} \\ P o s i t i o n v e c t o r o f C = 2 \hat{i} - 3 \hat{k} \\ P o s i t i o n v e c t o r o f D = 3 \hat{i} - 2 \hat{j} + \hat{k} \\ \vec{A B} = P . V o f B - P . V o f A \\ = (2 \hat{i} - \hat{j} + 3 \hat{k}) - (\hat{i} + \hat{j} - \hat{k}) = \hat{i} - 2 \hat{j} + 4 \hat{k} \\ \vec{C D} = P . V o f D - P . V o f C \\ = (3 \hat{i} - 2 \hat{j} + \hat{k}) - (2 \hat{i} - 3 \hat{k}) = \hat{i} - 2 \hat{j} + 4 \hat{k} \\ Projection o f \vec{A B} o n \vec{C D} = \frac{\bar{A B} \bar{. C D}}{| \bar{C D} |} = \frac{(\hat{i} - 2 \hat{j} + 4 \hat{k}) (\hat{i} - 2 \hat{j} + 4 \hat{k})}{\sqrt[]{{(1)}^{2} + {(- 2)}^{2} + {(4)}^{2}}} \\ = \frac{1 + 4 + 16}{\sqrt[]{1 + 4 + 16}} = \frac{21}{\sqrt[]{21}} = \sqrt[]{21} \\ H e n c e, t h e r e q u i r e d p rojection= \sqrt[]{21} . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Find the sine of the angle between the vectors $\vec{a}$ = 3î + 2ĵ + k̂ and $\vec{b}$ = −î + 2ĵ + 4k̂.

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n t h a t \vec{a} = 3 \hat{i} + \hat{j} + 2 \hat{k} a n d \vec{b} = 2 \hat{i} - 2 \hat{j} + 4 \hat{k} \\ W e k n o w t h a t | \vec{a} \times \vec{b} | = | \vec{a} | | \vec{b} | s i n θ \\ ∴ \vec{a} \times \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 2 & - 2 & 4 \end{matrix} | \\ = \hat{i} (4 + 4) - \hat{j} (12 - 4) + \hat{k} (- 6 - 2) \\ = 8 \hat{i} - 8 \hat{j} - 8 \hat{k} \\ ∴ | \vec{a} \times \vec{b} | = \sqrt[]{{(8)}^{2} + {(- 8)}^{2} + {(- 8)}^{2}} \\ = \sqrt[]{64 + 64 + 64} = \sqrt[]{192} = \sqrt[]{64 \times 3} = 8 \sqrt[]{3} \\ \Rightarrow | \vec{a} | = \sqrt[]{{(3)}^{2} + {(1)}^{2} + {(2)}^{2}} = \sqrt[]{9 + 1 + 4} = \sqrt[]{14} \\ \Rightarrow | \vec{b} | = \sqrt[]{{(2)}^{2} + {(- 2)}^{2} + {(4)}^{2}} = \sqrt[]{4 + 4 + 16} = \sqrt[]{24} = 2 \sqrt[]{6} \\ ∴ s i n θ = \frac{| \vec{a} \times \vec{b} |}{| \vec{a} | | \vec{b} |} = \frac{8 \sqrt[]{3}}{\sqrt[]{14} . 2 \sqrt[]{6}} = \frac{4 \sqrt[]{3}}{\sqrt[]{84}} = \frac{4 \sqrt[]{3}}{2 \sqrt[]{21}} = \frac{2}{\sqrt[]{7}} \\ H e n c e, s i n θ = \frac{2}{\sqrt[]{7}} . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If $\vec{a}$ + $\vec{b}$ + $\vec{c}$ = 0, show that $\vec{a}$ × $\vec{b}$ = $\vec{b}$ × $c$ = $\vec{c}$ × $\vec{a}$ . Interpret the result geometrically.

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n t h a t \vec{a} + \vec{b} + \vec{c} = 0 \\ S o, \vec{a} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{a} \times 0 \\ \Rightarrow \vec{a} \times \vec{a} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = 0 \\ \Rightarrow \vec{0} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = 0 (? \vec{a} \times \vec{a} = 0) \\ \Rightarrow \vec{a} \times \vec{b} - \vec{c} \times \vec{a} = 0 (\vec{a} \times \vec{c} = - \vec{c} \times \vec{a}) \\ \Rightarrow \vec{a} \times \vec{b} = \vec{c} \times \vec{a} \dots (i) \\ N o w, \vec{a} + \vec{b} + \vec{c} = 0 \\ \Rightarrow \vec{b} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{b} \times 0 \\ \Rightarrow \vec{b} \times \vec{a} + \vec{b} \times \vec{b} + \vec{b} \times \vec{c} = 0 \\ \Rightarrow \vec{b} \times \vec{a} + \vec{0} + \vec{b} \times \vec{c} = 0 (? \vec{b} \times \vec{b} = 0) \\ \Rightarrow - (\vec{a} \times \vec{b}) + \vec{b} \times \vec{c} = 0 \\ \Rightarrow \vec{b} \times \vec{c} = \vec{a} \times \vec{b} \dots (i i) \\ F r o m e q . (i) a n d (i i) w e g e t \\ \vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} . H e n c e p r o v e d . \\ G e o m e t r i c a l I n t e r p r e t a t i o n \\ A c c o r d i n g t o f i g u r e, w e h a v e A r e a o f parallelogram A B C D i s \\ \Rightarrow | \vec{a} \times \vec{b} | = | \vec{a} | | \vec{b} | s i n θ \\ Since, t h e parallelograms o n t h e s a m e b a s e a n d b e t w e e n t h e s a m e p a r a l l e l l i n e s a r e \\ e q u a l i n a r e a . \\ ∴ | \vec{a} \times \vec{b} | = | \vec{b} \times \vec{c} | = | \vec{c} \times \vec{a} | \\ \Rightarrow \vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Find the angle between the vectors 2î − ĵ + k̂ and 3î + 4ĵ − k̂.

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} L e t \vec{a} = 2 \hat{i} - \hat{j} + \hat{k} a n d \vec{b} = 3 \hat{i} + 4 \hat{j} - \hat{k} \\ a n d l e t θ b e t h e a n g l e b e t w e e n \vec{a} a n d \vec{b} . \\ ∴ c o s θ = \frac{\vec{a} . \vec{b}}{| \vec{a} | | \vec{b} |} = \frac{(2 \hat{i} - \hat{j} + \hat{k}) (3 \hat{i} + 4 \hat{j} - \hat{k})}{\sqrt[]{4 + 1 + 1} . \sqrt[]{9 + 16 + 1}} \\ = \frac{6 - 4 - 1}{\sqrt[]{6} \sqrt[]{26}} = \frac{1}{2 \sqrt[]{3} . \sqrt[]{13}} = \frac{1}{2 \sqrt[]{39}} \\ ∴ θ = c o s^{- 1} \frac{1}{2 \sqrt[]{39}} \Rightarrow c o s^{- 1} (\frac{1}{156}) \\ H e n c e, t h e r e q u i r e d v a l u e o f θ i s c o s^{- 1} (\frac{1}{156}) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Find a vector of magnitude 6, which is perpendicular to both the vectors 2î − 2ĵ + k̂ and 4î − 3ĵ + k̂.

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} L e t \vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k} a n d \vec{b} = 4 \hat{i} - \hat{j} + 3 \hat{k} \\ W e k n o w t h a t u n i t v e c t o r a r e p e r p e n d i c u l a r t o \vec{a} a n d \vec{b} = \frac{(\vec{a} \times \vec{b})}{| \vec{a} \times \vec{b} |} \\ \vec{a} \times \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 1 & 2 \\ 4 & - 1 & 3 \end{matrix} | \\ = \hat{i} (- 3 + 2) - \hat{j} (6 - 8) + \hat{k} (- 2 + 4) = - \hat{i} + 2 \hat{j} + 2 \hat{k} \\ ∴ | \vec{a} \times \vec{b} | = \sqrt[]{{(- 1)}^{2} + {(2)}^{2} + {(2)}^{2}} = \sqrt[]{1 + 4 + 4} = \sqrt[]{9} = 3 \\ S o, \frac{(\vec{a} \times \vec{b})}{| \vec{a} \times \vec{b} |} = \frac{- \hat{i} + 2 \hat{j} + 2 \hat{k}}{3} = \frac{1}{3} (- \hat{i} + 2 \hat{j} + 2 \hat{k}) \\ N o w t h e v e c t o r o f m a g n i t u d e 6 = \frac{1}{3} (- \hat{i} + 2 \hat{j} + 2 \hat{k}) . 6 \\ = 2 (- \hat{i} + 2 \hat{j} + 2 \hat{k}) = - 2 \hat{i} + 4 \hat{j} + 4 \hat{k} \\ H e n c e, t h e r e q u i r e d v e c t o r i s - 2 \hat{i} + 4 \hat{j} + 4 \hat{k} . \end{array}$

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Show that the area of the parallelogram whose diagonals are given by $\vec{a}$ and $\vec{b}$ is $\frac{1}{2} ∣ \frac{\vec{a} \times \vec{b}}{2} ∣$ Also, find the area of the parallelogram whose diagonals are 2î − ĵ + k̂ and î + 3ĵ − k̂.

Prove that in any triangle ABC, $c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$ where a, b, c are the magnitudes of the sides opposite to the vertices A, B, C, respectively.

Using vectors, prove that parallelograms on the same base and between the same parallels are equal in area.

Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, –1, 4), C(4, 5, –1).

If A, B, C, D are the points with position vectors î + ĵ − k̂, 2î − 3ĵ + k̂, 2î − 3k̂, and 3î + 2ĵ − k̂, respectively, find the projection of $\bar{A B}$ along $\bar{C D .}$

Find the sine of the angle between the vectors $\vec{a}$ = 3î + 2ĵ + k̂ and $\vec{b}$ = −î + 2ĵ + 4k̂.

If $\vec{a}$ + $\vec{b}$ + $\vec{c}$ = 0, show that $\vec{a}$ × $\vec{b}$ = $\vec{b}$ × $c$ = $\vec{c}$ × $\vec{a}$ . Interpret the result geometrically.

Find the angle between the vectors 2î − ĵ + k̂ and 3î + 4ĵ − k̂.

Find a vector of magnitude 6, which is perpendicular to both the vectors 2î − 2ĵ + k̂ and 4î − 3ĵ + k̂.