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Question

Mathematics Question on Vector Algebra

If a=2,b=7|\vec{a} | = 2 , |\vec{b}| = 7 and a×b=3i^+2j^+6k^\vec{a} \times \vec{b} = 3 \hat{i} + 2 \hat{j} + 6\hat{k} then the angle between a\vec{a} and b\vec{b} is

A

π3\frac{\pi}{3}

B

π6\frac{\pi}{6}

C

π2\frac{\pi}{2}

D

π4\frac{\pi}{4}

Answer

π6\frac{\pi}{6}

Explanation

Solution

a=2,b=7|\vec{a} | = 2 , |\vec{b}| = 7
a×b=3i^+2j^+6k^\vec{a} \times \vec{b} = 3 \hat{i} + 2 \hat{j} + 6\hat{k}
we know that a×b=absinθn^\vec{a} \times \vec{b} = |\vec{a} ||\vec{b}| \sin \theta \hat{n}
or a×b=absinθ.n^| \vec{a} \times \vec{b} | = | |\vec{a}||\vec{b}| \sin \, \theta . \hat{n}|
9+4+36=absinθ.1\Rightarrow \: \sqrt{9 + 4 + 36} = | \vec{a} ||\vec{b}||\sin \, \theta| .1
49=2×7sinθ\Rightarrow \: \sqrt{49} = 2 \times 7 |\sin \, \theta|
12=sinθθ=π6\Rightarrow\:\: \frac{1}{2} = \sin \theta \Rightarrow \:\:\: \theta = \frac{\pi}{6}