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Solved Board Paper

Q1
Marks:1
Answer:

Explanation:

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Q2
$L e t \vec{A} = 2 \hat{i} + \hat{k}, B = \hat{i} + \hat{j} + \hat{k} a n d C = 4 \hat{i} - 3 \hat{j} + 7 \hat{k} .$

$T h e v e c t o r \vec{R} w h i c h s a t i s f i e s t h e e q u a t i o n s \vec{R} \times \vec{B} = \vec{C} \times \vec{B} a n d \vec{R} . \vec{A} = 0 i s g i v e n b y$
Marks:1
Answer:

$- \hat{i} - 8 \overset{⏜}{j} + 2 \hat{k}$

Explanation:

$L e t \vec{R} = x \hat{i} + y \hat{j} + z \hat{k} . T h e n$

$\vec{R} \times \vec{B} = \vec{C} \times \vec{B} \Rightarrow (\vec{R} - \vec{C}) \times \vec{B} = 0$

$\Rightarrow |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ x - 4 & y + 3 & z - 7 \\ 1 & 1 & 1 \end{matrix}| = \vec{0}$

$\Rightarrow (y - z + 10) \hat{i} + (z - x - 3) \hat{j} + (x - y - 7) \hat{k} = \vec{0}$

$\Rightarrow y - z = 10, z - x = 3, x - y = 7$

$A l s o, \vec{R} . \vec{A} = 0 \Rightarrow 2 x + 0 . y + z = 0 \Rightarrow z = - 2 x$

On solving, we obtain

x = –1, y = –8, z = 2

$H e n c e, \vec{R} = - \hat{i} - 8 \hat{j} + 2 \hat{k}$

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Q3
Marks:1
Answer:

Explanation:

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Q4
A unit vector perpendicular to the plane ABC, where A, B and C
are respectively the points (3, –1, 2), (1, –1, –3) and (4, –3, 1), is
Marks:1
Answer:

Explanation:

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Q5
Marks:1
Answer:

Explanation:

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