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New answer posted

a year ago

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

If $\vec{| a} . * \vec{b} |^{2} + \vec{| a} . \vec{b} |^{2} = 144 a n d \vec{| a} | = 4,$ then b 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:

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

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New answer posted

a year ago

Class 12th 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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Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

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

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

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 12th 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 ________.

0 Follower 7 Views

alok kumar singh

Contributor-Level 10

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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New answer posted

a year ago

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

Find the equations of the line passing through the point (3, 0, 1) and parallel to the planes $x + 2 y = 0$ and $3 y - z = 0$ .

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n point i s (3, 0, 1) a n d t h e e q u a t i o n o f p l a n e s a r e \\ x + 2 y = 0 \dots (i) \\ a n d 3 y - z = 0 \dots (i i) \\ E q u a t i o n o f a n y l i n e l passing t h r o u g h (3, 0, 1) i s \\ l : \frac{x - 3}{a} = \frac{y - 0}{b} = \frac{z - 1}{c} \\ D i r e c t i o n r a t i o s o f t h e n o r m a l t o t h e p l a n e (i) a n d (i i) a r e (1, 2, 0) a n d (0, 3, - 1) \\ Since, t h e l i n e i s p a r a l l e l t o b o t h t h e p l a n e s . \\ ∴ 1 . a + 2 . b + 0 . c = 0 \Rightarrow a + 2 b + 0 c = 0 \\ a n d 0 . a + 3 . b - 1 . c = 0 \Rightarrow 0 . a + 3 b - c = 0 \\ S o, \frac{a}{- 2 - 0} = \frac{- b}{- 1 - 0} = \frac{c}{3 - 0} = λ \\ ∴ a = - 2 λ, b = λ, c = 3 λ \\ S o, e q u a t i o n o f l i n e i s \\ \frac{x - 3}{- 2 λ} = \frac{y - 0}{λ} = \frac{z - 1}{3 λ} \\ H e n c e, t h e r e q u i r e d e q u a t i o n \frac{x - 3}{- 2} + \frac{y - 0}{1} + \frac{z - 1}{3} \\ o r i n v e c t o r f o r m i s (x - 3) \hat{i} + y \hat{j} + (z - 1) \hat{k} = λ (- 2 \hat{i} + \hat{j} + 3 \hat{k}) \end{array}$

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New answer posted

a year ago

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

The vectors $\vec{a}$ = 3î − 2ĵ + 2k̂ and b̂ = î − 2k̂ are the adjacent sides of a parallelogram. The acute angle between its diagonals is ________.

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t \vec{a} = 3 \hat{i} - 2 \hat{j} + 2 \hat{k} \\ a n d \vec{b} = - \hat{i} - 2 \hat{k} \\ ∴ \vec{a} + \vec{b} = 2 \hat{i} - 2 \hat{j} a n d \vec{a} - \vec{b} = 4 \hat{i} - 2 \hat{j} + 4 \hat{k} \\ L e t θ b e t h e a n g l e b e t w e e n t h e t w o d i a g o n a l v e c t o r s \vec{a} + \vec{b} a n d \vec{a} - \vec{b} t h e n \\ c o s θ = \frac{(\vec{a} + \vec{b}) . (\vec{a} - \vec{b})}{| \vec{a} + \vec{b} | | \vec{a} - \vec{b} |} = \frac{(2 \hat{i} - 2 \hat{j}) . (4 \hat{i} - 2 \hat{j} + 4 \hat{k})}{\sqrt[]{{(2)}^{2} + {(- 2)}^{2}} \sqrt[]{{(4)}^{2} + {(- 2)}^{2} + {(4)}^{2}}} \\ = \frac{8 + 4}{2 \sqrt[]{2} . 6} = \frac{12}{2 \sqrt[]{2} . 6} = \frac{1}{\sqrt[]{2}} \\ ∴ θ = \frac{π}{4} \\ H e n c e, t h e v a l u e o f r e q u i r e d f i l l e r i s \frac{π}{4} . \end{array}$

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New answer posted

a year ago

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

Find the length and the foot of the perpendicular from the point ( $1, \frac{3}{2}, 2$ ) to the plane $2 x - 2 y + 4 z + 5 = 0$ .

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n p l a n e i s 2 x - 2 y + 4 z + 5 = 0 a n d t h e g i v e n points i s (1, \frac{3}{2}, 2) \\ D' r a t i o s o f t h e n o r m a l t o t h e p l a n e a r e 2, - 2, 4 \\ S o, t h e ? e q u a t i o n o f l i n e passing t h r o u g h (1, \frac{3}{2}, 2) a n d w h o s e d' r a t i o s a r e e q u a l t o t h e \\ D' r a t i o s o f t h e n o r m a l t o t h e p l a n e i . e ., 2, - 2, 4 i s \frac{x - 1}{2} = \frac{y - \frac{3}{2}}{- 2} = \frac{z - 2}{4} = λ \\ ∴ A n y point i n t h e p l a n e i s 2 λ + 1, - 2 λ + \frac{3}{2}, 4 λ + 2 \\ Since, t h e point l i e s i n t h e p l a n e, t h e n \\ 2 (2 λ + 1) - 2 (- 2 λ + \frac{3}{2}) + 4 (4 λ + 2) + 5 = 0 \\ \Rightarrow 4 λ + 2 + 4 λ - 3 + 16 λ + 8 + 5 = 0 \\ \Rightarrow 24 λ + 12 = 0 ∴ λ = - \frac{1}{2} \\ S o, t h e c o o r d i n a t e s o f t h e point i n t h e p l a n e a r e \\ 2 (- \frac{1}{2}) + 1, - 2 (- \frac{1}{2}) + \frac{3}{2}, 4 (- \frac{1}{2}) + 2 i . e ., 0, \frac{5}{2}, 0 \\ H e n c e, t h e f o o t o f t h e p e r p e n d i c u l a r i s (0, \frac{5}{2}, 0) a n d t h e \\ r e q u i r e d l e n g t h = \sqrt[]{{(1 - 0)}^{2} + {(\frac{3}{2} - \frac{5}{2})}^{2} + {(2 - 0)}^{2}} \\ = \sqrt[]{1 + 1 + 4} = \sqrt[]{6} u n i t s \end{array}$

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New answer posted

a year ago

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

Find the distance of the point (2, 4, –1) from the line $\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{– 9} .$

0 Follower 17 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} T h e g i v e n e q u a t i o n o f l i n e i s \\ \frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{– 9} = λ a n d a n y point P (2, 4, - 1) \\ L e t Q b e a n y point o n t h e g i v e n l i n e \\ ∴ C o o r d i n a t e s o f Q a r e x = λ - 5, y = 4 λ - 3 a n d z = - 9 λ + 6 \\ a n d t h e g i v e n point i s P (2, 3, - 8) \\ D i r e c t i o n r a t i o s o f P Q a r e λ - 5 - 2, 4 λ - 3 - 4, - 9 λ + 6 + 1 \\ i . e ., λ - 7, 4 λ - 7, - 9 λ + 7 \\ a n d t h e D' r a t i o s o f t h e g i v e n l i n e a r e 1, 4, - 9 . \\ I f P Q ⊥ l i n e \\ t h e n 1 (λ - 7) + 4 (4 λ - 7) - 9 (- 9 λ + 7) = 0 \\ \Rightarrow λ - 7 + 16 λ - 28 + 81 λ - 63 = 0 \\ \Rightarrow 98 λ - 98 = 0 \Rightarrow λ = 1 \\ S o, C o o r d i n a t e s o f Q a r e 1 - 5, 4 * 1 - 3, - 9 * 1 + 6 \\ i . e ., - 4, 1, - 3 \\ N o w, distance P Q = \sqrt[]{{(- 4 - 2)}^{2} + {(1 - 4)}^{2} + {(- 3 + 1)}^{2}} \\ = \sqrt[]{{(- 6)}^{2} + {(- 3)}^{2} + {(- 2)}^{2}} = \sqrt[]{36 + 9 + 4} = \sqrt[]{49} = 7 \\ H e n c e, t h e r e q u i r e d distance 7 u n i t s . \end{array}$

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New question posted

a year ago

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

Find the distance of the point (2, 4, –1) from the line $\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .$

0 Follower 4 Views

New answer posted

a year ago

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

If $\vec{r}$ . $\vec{a}$ = 0, $\vec{r}$ . $\vec{b}$ = 0, $\vec{r}$ . $\vec{c}$ = 0 for some non-zero vector r̂, then the value of
( $\vec{a}$ * $\vec{b}$ ) ĉ is ________.

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} I f \vec{r} i s a n o n - z e r o v e c t o r, t h e n \vec{a}, \vec{b} a n d \vec{c} c a n b e i n t h e s a m e p l a n e . \\ Since a n g l e s b e t w e e n \vec{a}, \vec{b} a n d \vec{c} a r e z e r o i . e . θ = 0 \\ ∴ \vec{a} . (\vec{b} * \vec{c}) = 0 \\ H e n c e, t h e r e q u i r e d v a l u e i s 0 . \end{array}$

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