Two vectors a and b are perpendicular to each other- A vector c is inclined at an angle - to both a and b- If -a-2-b-3- -c-2 and c-pa-qb-ra-b- then

Here nbsp;|a|=2,|b|=3 and nbsp;|c|=2 ⇒cosθ=a.c4=b.c6 c=pa+qb+r(a×b) ⋯(1) Taking dot product with a on both sides in equation (1) nbsp; ⇒a.c=p(a.a)+q( ...

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Byju's Answer

Standard XII

Mathematics

Linear Dependence and Independence of Vectors

Two vectors a...

Question

Two vectors $\to a$ and $\to b$ are perpendicular to each other. A vector $\to c$ is inclined at an angle $θ$ to both $\to a$ and $\to b$ . If $| \to a | = 2$ , $| \to b | = 3$ , $| \to c | = 2$ and $\to c = p \to a + q \to b + r (\to a \times \to b)$ , then

p^{2} = {cos}^{2} θ

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r^{2} = \frac{- cos 2 θ}{9}

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r^{2} = \frac{- {cos}^{2} θ}{9}

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q^{2} = \frac{4 {cos}^{2} θ}{9}

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Solution

The correct option is D $q^{2} = \frac{4 {cos}^{2} θ}{9}$
Here $| \to a | = 2, | \to b | = 3$ and $| \to c | = 2$
$\Rightarrow cos θ = \frac{\to a . \to c}{4} = \frac{\to b . \to c}{6}$
$\to c = p \to a + q \to b + r (\to a \times \to b) \dots (1)$

Taking dot product with $\to a$ on both sides in equation $(1)$
$\Rightarrow \to a . \to c = p (\to a . \to a) + q (\to a . \to b) + r \to a . (\to a \times \to b)$
$\Rightarrow 4 cos θ = 4 p$ $\Rightarrow cos θ = p$

Similarly taking dot product with $\to b$ on both sides in equation $(1)$
$\Rightarrow \to b . \to c = p (\to a . \to b) + q (\to b . \to b) + r \to b . (\to a \times \to b)$
$\Rightarrow 6 cos θ = 9 q$
$\Rightarrow q = \frac{2}{3} p$

Squaring eqn $(1),$ we get
$| \to c |^{2} = p^{2} | \to a |^{2} + q^{2} | \to b |^{2} + r^{2} | (\to a \times \to b) |^{2} + 2 p q (\to a . \to b) + 2 p r (\to a) . (\to a \times \to b) + 2 q r (\to b) . (\to a \times \to b)$

$\Rightarrow 4 = 4 p^{2} + 9 q^{2} + r^{2} | (\to a \times \to b) |^{2}$
$\Rightarrow 4 = 8 p^{2} + r^{2} | \to a |^{2} ∥ \to b |^{2} \times 1$
$\Rightarrow 4 = 8 p^{2} + 36 r^{2}$

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