A and b are two vectors such that - a -1- b -4 and a - b -2- If c -2 a - b -3 b- then the angle between b and C is3 -A- 5 -5 -6C- 3D- 1-6

The correct option is B 5π6|a|=1,|b|=4,a·b =2, c=(2a×b) 3b ⇒c+3b=2a×b ∵a·b=2 ⇒cosθ =24=12 ⇒θ=π3 ⇒ |c+3b|^2=|2a×b|^2 Taking b·c=|b||c|cosα ⇒ |c|^2 +144+6b·c=48 ...

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Byju's Answer

Standard XII

Mathematics

Cross Product of Two Vectors

a and b are t...

Question

$\to a$ and $\to b$ are two vectors such that $| \to a | = 1, | \to b | = 4$ and $\to a \cdot \to b = 2$ . If $\to c = (2 \to a \times \to b) - 3 \to b$ , then the angle between $\to b$ and $\to c$ is

\frac{π}{3}

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\frac{π}{6}

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\frac{3 π}{4}

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\frac{5 π}{6}

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Solution

The correct option is D $\frac{5 π}{6}$
$| \to a | = 1, | \to b | = 4, \to a \cdot \to b = 2, \to c = (2 \to a \times \to b) - 3 \to b$
$\Rightarrow \to c + 3 \to b = 2 \to a \times \to b ∵ \to a \cdot \to b = 2 \Rightarrow cos θ = \frac{2}{4} = \frac{1}{2} \Rightarrow θ = \frac{π}{3} \Rightarrow | \to c + 3 \to b |^{2} = | 2 \to a \times \to b |^{2}$
Taking $\to b \cdot \to c = | \to b | | \to c | cos α$
$\Rightarrow | \to c |^{2} + 144 + 6 \to b \cdot \to c = 48 \Rightarrow | \to c |^{2} + 96 + 6 (\to b \cdot \to c) = 0$

$\to c = (2 \to a \times \to b) - 3 \to b$
$\Rightarrow \to b \cdot \to c = 0 - 3 \times 16 = - 48$

$\Rightarrow | \to c |^{2} + 96 - 6 \times 48 = 0 \Rightarrow | \to c |^{2} = 192$
Taking $\to b \cdot \to c = | \to b | | \to c | cos α$
$\Rightarrow 192 + 96 + 6 | \to b | \cdot | \to c | cos α = 0 \Rightarrow cos α = - \frac{288}{6 \times 4 \times 8 \sqrt{3}} = - \frac{3}{2 \sqrt{3}} = - \frac{\sqrt{3}}{2} \Rightarrow α = \frac{5 π}{6}$

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→a and →b are two vectors such that |→a|=1,|→b|=4 and →a⋅→b=2. If →c=(2→a×→b)−3→b, then the angle between →b and →c is

$\to a$ and $\to b$ are two vectors such that $| \to a | = 1, | \to b | = 4$ and $\to a \cdot \to b = 2$ . If $\to c = (2 \to a \times \to b) - 3 \to b$ , then the angle between $\to b$ and $\to c$ is