If 3 non zero vectors a- - b- - c- are such that a-b- - 2 a-c- - - a- - - - c - - 1 - - b- - - 4 the angle between b- and c- is cos - 1 1 - 4 then b- - -c- - -a- where - - - - - - - is -

The correct option is B 6 We have, a⃗×( b⃗ 2c⃗ #160; ) =0⃗ #160; ⇒b⃗ 2c⃗ =μa⃗ #160; | b⃗ 2c⃗ #160; | =| μ #160; | #160; ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Complex Numbers

If 3 non ze...

Question

If $3$ non zero vectors $\to a, \to b, \to c$ are such that $\to a \times \to b = 2 (\to a \times \to c), | \to a | = | ¯ ¯ c | = 1; | \to b | = 4$ the angle between $\to b and \to c is {cos}^{- 1} \frac{1}{4} then \to b = ℓ \to c + μ \to a where | ℓ | + | μ | is$ -

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Solution

The correct option is B $6$
We have,
$\to a \times (\to b - 2 \to c) = \to 0 \Rightarrow \to b - 2 \to c = μ \to a ∣ ∣ \to b - 2 \to c ∣ ∣ = | μ | {∣ ∣ \to b - 2 \to c ∣ ∣}^{2} = {| μ |}^{2} (\to b - 2 \to c) . (\to b - 2 \to c) = {| μ |}^{2} 16 + 4 - 4 \times 4 \times 1 \times \frac{1}{4} = μ^{2} 16 = μ^{2} μ = 4$

Since,
$\to b = μ \to a + 2 \to c = 4 + 2 = 6$

$H e n c e, t h e o p t i o n A i s t h e c o r r e c t .$

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If 3 non zero vectors →a,→b,→c are such that →a×→b=2(→a×→c),|→a|=|¯¯c|=1;|→b|=4 the angle between →b and →c is cos−114 then →b=ℓ→c+μ→a where |ℓ|+|μ| is -

The correct option is B 6We have,→a×(→b−2→c)=→0⇒→b−2→c=μ→a∣∣→b−2→c∣∣=|μ|∣∣→b−2→c∣∣2=|μ|2(→b−2→c).(→b−2→c)=|μ|216+4−4×4×1×14=μ216=μ2μ=4Since,→b=μ→a+2→c=4+2=6Hence,theoptionAisthecorrect.

If $3$ non zero vectors $\to a, \to b, \to c$ are such that $\to a \times \to b = 2 (\to a \times \to c), | \to a | = | ¯ ¯ c | = 1; | \to b | = 4$ the angle between $\to b and \to c is {cos}^{- 1} \frac{1}{4} then \to b = ℓ \to c + μ \to a where | ℓ | + | μ | is$ -