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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If $\vec{| a}$ + $\vec{b} |$ = $\vec{| a}$ - $\vec{b} |$ , then the vectors $\vec{a}$ and $\vec{b}$ are orthogonal.

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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 True or False type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t | \vec{a} + \vec{b} | = | \vec{a} - \vec{b} | \\ S q u a r i n g b o t h s i d e s, w e g e t \\ {| \vec{a} + \vec{b} |}^{2} = {| \vec{a} - \vec{b} |}^{2} \\ \Rightarrow {| \vec{a} |}^{2} + {| \vec{b} |}^{2} + 2 \vec{a} . \vec{b} = {| \vec{a} |}^{2} + {| \vec{b} |}^{2} - 2 \vec{a} . \vec{b} \\ \Rightarrow 2 \vec{a} . \vec{b} = - 2 \vec{a} . \vec{b} \\ \Rightarrow \vec{a} . \vec{b} = - \vec{a} . \vec{b} \\ \Rightarrow 2 \vec{a} . \vec{b} = 0 \Rightarrow \vec{a} . \vec{b} = 0 \\ w h i c h i m p l i e s t h a t \vec{a} a n d \vec{b} a r e o r t h o g o n a l . \\ H e n c e t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

The position vector of a point P is a vector whose initial point is the origin.

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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 True or False type Questions as classified in NCERT Exemplar

Sol: $T r u e .$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Find the equation of the plane which is perpendicular to the plane  $5 x + 3 y + 6 z + 8 = 0$ and contains the line of intersection of the planes  $x + 2 y + 3 z – 4 = 0$ and $2 x + y – z + 5 = 0$ .

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 long answer type question as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n p l a n e s a r e \\ P_{1} : 5 x + 3 y + 6 z + 8 = 0 \\ P_{2} : x + 2 y + 3 z - 4 = 0 \\ P_{3} : 2 x + y - z + 5 = 0 \\ E q u a t i o n s o f t h e p l a n e passing t h r o u g h t h e l i n e o f intersection o f P_{2} a n d P_{3} i s \\ (x + 2 y + 3 z - 4) + λ (2 x + y - z + 5) = 0 \\ \Rightarrow (1 + 2 λ) x + (2 + λ) y + (3 - λ) z - 4 + 5 λ = 0 \dots (i) \\ P l a n e (i) i s p e r p e n d i c u l a r t o P_{1}, t h e n \\ 5 (1 + 2 λ) + 3 (2 + λ) + 6 (3 - λ) = 0 \\ \Rightarrow 5 + 10 λ + 6 + 3 λ + 18 - 6 λ = 0 \\ \Rightarrow 7 λ + 29 = 0 \Rightarrow λ = \frac{- 29}{7} \\ P u t t i n g t h e v a l u e o f λ i n e q . (i), w e g e t \\ [1 + 2 (\frac{- 29}{7})] x + [2 - \frac{29}{7}] y + [3 + \frac{29}{7}] z - 4 + 5 (\frac{- 29}{7}) = 0 \\ \Rightarrow \frac{- 15}{7} x - \frac{15}{7} y + \frac{50}{7} z - 4 - \frac{145}{7} = 0 \\ \Rightarrow - 15 x - 15 y + 50 z - 28 - 145 = 0 \\ \Rightarrow - 15 x - 15 y + 50 z - 173 = 0 \Rightarrow 15 x + 15 y - 50 z + 173 = 0 \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

State True or False for the statement in each of the Exercises 41 to 45.

If $\vec{| a}$ | =| $\vec{b} |$ , then necessarily it implies $\vec{a}$ = ± $\vec{b}$

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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 True or False type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} I f | \vec{a} | = | \vec{b} | t h e n \vec{a} = \pm \vec{b} w h i c h i s t r u e . \\ H e n c e t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Find the shortest distance between the lines $\vec{r} = (8 + 3 λ) \hat{i} - (9 + 16 λ) \hat{j} + (10 + 7 λ) \hat{k}$ and $\vec{r} = 15 \hat{i} - 29 \hat{j} + 5 \hat{k} + μ (3 \hat{i} + 8 \hat{j} - 5 \hat{k})$

3 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n e q u a t i o n s o f l i n e s a r e \\ \vec{r} = (8 + 3 λ) \hat{i} - (9 + 16 λ) \hat{j} + (10 + 7 λ) \hat{k} \dots (i) \\ a n d \vec{r} = 15 \hat{i} - 29 \hat{j} + 5 \hat{k} + μ (3 \hat{i} + 8 \hat{j} - 5 \hat{k}) \dots (i i) \\ E q u a t i o n (i) c a n b e r e - w r i t t e n a s \\ \vec{r} = 8 \hat{i} - 9 \hat{j} + 10 \hat{k} + λ (3 \hat{i} - 16 \hat{j} + 7 \hat{k}) \dots (i i i) \\ H e r e, \vec{a_{1}} = 8 \hat{i} - 9 \hat{j} + 10 \hat{k} a n d \vec{a_{2}} = 15 \hat{i} - 29 \hat{j} + 5 \hat{k} \\ \vec{b_{1}} = 3 \hat{i} - 16 \hat{j} + 7 \hat{k} a n d \vec{b_{2}} = 3 \hat{i} + 8 \hat{j} - 5 \hat{k} \\ \vec{a_{2}} - \vec{a_{1}} = 7 \hat{i} + 38 \hat{j} - 5 \hat{k} \\ \vec{b_{1}} \times \vec{b_{2}} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 16 & 7 \\ 3 & 8 & - 5 \end{matrix} | \\ = \hat{i} (80 - 56) - \hat{j} (- 15 - 21) + \hat{k} (24 + 48) \\ = 24 \hat{i} + 36 \hat{j} + 72 \hat{k} \\ ∴ S h o r t e s t distance, S D = | \frac{(\vec{a_{2}} - \vec{a_{1}}) . (\vec{b_{1}} \times \vec{b_{2}})}{| \vec{b_{1}} \times \vec{b_{2}} |} | = | \frac{(7 \hat{i} + 38 \hat{j} - 5 \hat{k}) . (24 \hat{i} + 36 \hat{j} + 72 \hat{k})}{\sqrt[]{{(24)}^{2} + {(36)}^{2} + {(72)}^{2}}} | \\ = | \frac{168 + 1368 - 360}{\sqrt[]{576 + 1296 + 5184}} | = | \frac{168 + 1008}{\sqrt[]{7056}} | = \frac{1176}{84} = 14 u n i t s \\ H e n c e, t h e r e q u i r e d distance i s 14 u n i t s . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If a is any non-zero vector, then $(\vec{| a} . \hat{i}) . \hat{i} + (\vec{a} . \hat{j} \cdot) \hat{j} + (\vec{a} . \hat{k}) \hat{k}$ equals _______.

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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 Fill in the blanks type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t \vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} \\ ∴ \vec{a} . \hat{i} = (a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}) . \hat{i} \\ = a_{1} \\ S i m i l a r l y, \vec{a} . \hat{j} = a_{2} a n d \vec{a} . \hat{k} = a_{3} \\ ∴ (\vec{a} . \hat{i}) . \hat{i} + (\vec{a} . \hat{j}) . \hat{j} + (\vec{a} . \hat{k}) . \hat{k} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} = \vec{a} \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s \vec{a} . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If $\vec{| a} . \times \vec{b} |^{2} + \vec{| a} . \vec{b} |^{2} = 144 a n d \vec{| a} | = 4,$ then b is equal to _______.

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

Sol:

$\begin{array}{l} {| \vec{a} \times \vec{b} |}^{2} + {(\vec{a} . \vec{b})}^{2} = 144 \\ \Rightarrow {(| \vec{a} | | \vec{b} | s i n θ)}^{2} + {(| \vec{a} | | \vec{b} | c o s θ)}^{2} = 144 \\ \Rightarrow {| \vec{a} |}^{2} {| \vec{b} |}^{2} s i n^{2} θ + {| \vec{a} |}^{2} {| \vec{b} |}^{2} c o s^{2} θ = 144 \\ \Rightarrow {| \vec{a} |}^{2} {| \vec{b} |}^{2} . (s i n^{2} θ + c o s^{2} θ) = 144 \\ \Rightarrow {| \vec{a} |}^{2} {| \vec{b} |}^{2} = 144 \\ \Rightarrow | \vec{a} | | \vec{b} | = 12 \\ \Rightarrow 4 . | \vec{b} | = 12 \\ ∴ | \vec{b} | = 3 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 3 . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Find the equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane $x - 2 y + 4 z = 10$ .

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Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} E q u a t i o n o f p l a n e passing t h r o u g h t w o points (x_{1}, y_{1,} z_{1}) a n d (x_{2}, y_{2}, z_{2}) w i t h i t s n o r m a l d' r a t i o s i s \\ a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0 \dots (i) \\ I f t h e p l a n e i s passing t h r o u g h t h e g i v e n points (2, 1, - 1) a n d (- 1, 3, 4) t h e n \\ a (x_{2} - x_{1}) + b (y_{2} - y_{1}) + c (z_{2} - z_{1}) = 0 \\ \Rightarrow a (- 1 - 2) + b (3 - 1) + c (4 + 1) = 0 \\ \Rightarrow - 3 a + 2 b + 5 c = 0 \dots (i i) \\ Since, t h e r e q u i r e d p l a n e i s p e r p e n d i c u l a r t o t h e g i v e n p l a n e x - 2 y + 4 z = 10, t h e n \\ 1 . a - 2 . b + 4 . c = 10 \dots (i i i) \\ S o l v i n g (i i) a n d (i i i) w e g e t, \\ \frac{a}{8 + 10} = \frac{- b}{- 12 - 5} = \frac{c}{6 - 2} = λ \\ a = 18 λ, b = 17 λ, c = 4 λ \\ H e n c e, t h e r e q u i r e d p l a n e i s 18 λ (x - 2) + 17 λ (y - 1) + 4 λ (z + 1) = 0 \\ \Rightarrow 18 x - 36 + 17 y - 17 + 4 z + 4 = 0 \\ \Rightarrow 18 x + 17 y + 4 z - 49 = 0 \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

The value of the expression $(\vec{a} \times b ?)^{2}$ + ${(\vec{a} . \vec{b})}^{2}$ is ________.

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 Fill in the blanks type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} {| \vec{a} \times \vec{b} |}^{2} + {(\vec{a} . \vec{b})}^{2} = {(| \vec{a} | | \vec{b} | s i n θ)}^{2} + {(| \vec{a} | | \vec{b} | c o s θ)}^{2} \\ = {| \vec{a} |}^{2} {| \vec{b} |}^{2} s i n^{2} θ + {| \vec{a} |}^{2} {| \vec{b} |}^{2} c o s^{2} θ \\ = {| \vec{a} |}^{2} {| \vec{b} |}^{2} . (s i n^{2} θ + c o s^{2} θ) \\ = {| \vec{a} |}^{2} {| \vec{b} |}^{2} . 1 = {| \vec{a} |}^{2} {| \vec{b} |}^{2} \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s {| \vec{a} |}^{2} {| \vec{b} |}^{2} . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Ten

The values of k for which |k $\vec{a}$ <| $\vec{a |}$ |  and k $\vec{a} + \frac{1}{2} \vec{a}$ is parallel to $\vec{a}$ holds true are ________.

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 Fill in the blanks type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t | k \vec{a} | < | \vec{a} | a n d k \vec{a} + \frac{1}{2} \vec{a} i s p a r a l l e l t o \vec{a} \\ ∴ | k \vec{a} | < | \vec{a} | \Rightarrow | k | | \vec{a} | < | \vec{a} | \Rightarrow | k | < 1 \Rightarrow - 1 < k < 1 \\ N o w since k \vec{a} + \frac{1}{2} \vec{a} i s p a r a l l e l t o \vec{a} \\ H e r e, w e s e e t h a t k = - \frac{1}{2}, k \vec{a} + \frac{1}{2} \vec{a} b e c o m e n u l l v e c t o r a n d t h e n i t w i l l n o t b e p a r a l l e l t o \vec{a} . \\ ∴ k \vec{a} + \frac{1}{2} \vec{a} i s p a r a l l e l t o \vec{a} w h e n k \in (- 1, 1) a n d k \neq \frac{1}{2} . \\ H e n c e, t h e r e q u i r e d v a l u e o f k \in (- 1, 1) a n d k \neq \frac{1}{2} . \end{array}$

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If $\vec{| a}$ + $\vec{b} |$ = $\vec{| a}$ - $\vec{b} |$ , then the vectors $\vec{a}$ and $\vec{b}$ are orthogonal.

The position vector of a point P is a vector whose initial point is the origin.

Find the equation of the plane which is perpendicular to the plane  $5 x + 3 y + 6 z + 8 = 0$ and contains the line of intersection of the planes  $x + 2 y + 3 z – 4 = 0$ and $2 x + y – z + 5 = 0$ .

State True or False for the statement in each of the Exercises 41 to 45.

If $\vec{| a}$ | =| $\vec{b} |$ , then necessarily it implies $\vec{a}$ = ± $\vec{b}$

Find the shortest distance between the lines $\vec{r} = (8 + 3 λ) \hat{i} - (9 + 16 λ) \hat{j} + (10 + 7 λ) \hat{k}$ and $\vec{r} = 15 \hat{i} - 29 \hat{j} + 5 \hat{k} + μ (3 \hat{i} + 8 \hat{j} - 5 \hat{k})$

If a is any non-zero vector, then $(\vec{| a} . \hat{i}) . \hat{i} + (\vec{a} . \hat{j} \cdot) \hat{j} + (\vec{a} . \hat{k}) \hat{k}$ equals _______.

If $\vec{| a} . \times \vec{b} |^{2} + \vec{| a} . \vec{b} |^{2} = 144 a n d \vec{| a} | = 4,$ then b is equal to _______.

Find the equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane $x - 2 y + 4 z = 10$ .

The value of the expression $(\vec{a} \times b ?)^{2}$ + ${(\vec{a} . \vec{b})}^{2}$ is ________.

The values of k for which |k $\vec{a}$ <| $\vec{a |}$ |  and k $\vec{a} + \frac{1}{2} \vec{a}$ is parallel to $\vec{a}$ holds true are ________.