Let a and b be two vectors such that -a-1- -b-4 and a-b-2- If c is a vector such that c- 2a-b-3b- then the angle between b and c is

Given, nbsp;|a|=1, |b|=4 and a·b=2. ⇒ |a×b|^2= |a|^2 |b|^2 (a·b)^2 ⇒ |a×b|^2=16 4=12 Now, nbsp;nbsp; c= nbsp;2a×b 3b nbsp; nbsp; ⇒ |c|^2 =4|a×b|^2 ...

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Byju's Answer

Standard XII

Physics

Position Vector

Let a and b b...

Question

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

\frac{π}{6}

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\frac{π}{3}

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\frac{2 π}{3}

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\frac{5 π}{6}

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Solution

The correct option is D $\frac{5 π}{6}$
Given, $| \to a | = 1,$ $| \to b | = 4$ and $\to a \cdot \to b = 2.$
$\Rightarrow | \to a \times \to b |^{2} = | \to a |^{2} | \to b |^{2} - (\to a \cdot \to b)^{2}$
$\Rightarrow | \to a \times \to b |^{2} = 16 - 4 = 12$

Now, $\to c = 2 \to a \times \to b - 3 \to b$
$\Rightarrow | \to c |^{2} = 4 | \to a \times \to b |^{2} + 9 | \to b |^{2}$
$\Rightarrow | \to c |^{2} = 4 \times 12 + 9 \times 16 = 192$

Let $θ$ be the angle between $\to b$ and $\to c .$
Then, $cos θ = \frac{\to b \cdot \to c}{| \to b | | \to c |}$
$\Rightarrow cos θ = \frac{- 48}{4 \times 8 \sqrt{3}} = - \frac{\sqrt{3}}{2}$
$\Rightarrow θ = \frac{5 π}{6}$

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

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