Three vectors a- b- c- are such that a-b-4a-c- and -a-b-1 and -c-1-4 - If the angle between b- and c- is -3- then b-is A- a -4 cB- a -4 cC- 4 c - aD- 2 c - a

The correct options are A a+4c nbsp; C 4c aa× nbsp;b=4(a× nbsp;c) ⇒a× nbsp;b=(a× 4c) ⇒a× nbsp;b (a× 4c)=0 ⇒a× (b 4c)=0 So, a and (b 4c) are collinear ...

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Byju's Answer

Standard XII

Mathematics

Three vectors...

Question

Three vectors $\to a, \to b, \to c$ are such that $\to a \times \to b = 4 (\to a \times \to c)$ and $| \to a | = | \to b | = 1$ and $| \to c | = \frac{1}{4}$ . If the angle between $\to b$ and $\to c$ is $\frac{π}{3}$ , then $\to b$ is

\to a + 4 \to c

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\to a - 4 \to c

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4 \to c - \to a

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2 \to c - \to a

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Solution

The correct options are
A $\to a + 4 \to c$
C $4 \to c - \to a$
$\to a \times \to b = 4 (\to a \times \to c)$
$\Rightarrow \to a \times \to b = (\to a \times 4 \to c)$
$\Rightarrow \to a \times \to b - (\to a \times 4 \to c) = 0$
$\Rightarrow \to a \times (\to b - 4 \to c) = 0$
So, $\to a$ and $(\to b - 4 \to c)$ are collinear or parallel vectors.
$\Rightarrow λ \to a = (\to b - 4 \to c) \dots [1]$
$\Rightarrow | λ \to a |^{2} = | \to b - 4 \to c |^{2}$
$\Rightarrow λ^{2} | \to a |^{2} = | \to b |^{2} + 16 | \to c |^{2} - 8 \times (\to b \cdot \to c)$
$\Rightarrow λ = \pm 1$

Putting $λ = 1$ in $[1]$ , we get
$\to b = 4 \to c + \to a$

Putting $λ = - 1$ in $[1]$ , we get
$\to b = 4 \to c - \to a$

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Similar questions

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Three vectors →a, →b, →c are such that →a×→b=4(→a×→c) and |→a|=|→b|=1 and |→c|=14. If the angle between →b and →c is π3, then →b is

Three vectors $\to a, \to b, \to c$ are such that $\to a \times \to b = 4 (\to a \times \to c)$ and $| \to a | = | \to b | = 1$ and $| \to c | = \frac{1}{4}$ . If the angle between $\to b$ and $\to c$ is $\frac{π}{3}$ , then $\to b$ is