Question

If vector A and vector B are two such vectors that Mod of A=2 & Mod of B=7 & A cross B= 3i cap+2j cap+6k cap, find angle between vector A and vector B

Answer 1

Answer:

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Answer 2

The angle between A and B is 30 degrees.

Explanation:

We have,

$|A|=2\\\\|B|=7\\\\A\times B=(3i+2j +6k)$

We need to find the angle between A and B. Using cross product formula to find it as :

$A\times B=|A||B|\sin \theta\\\\\sin \theta=\dfrac{|A\times B|}{|A||B|}\\\\\sin \theta=\dfrac{\sqrt{3^2+2^2+6^2} }{2\times 7}\\\\\sin\theta=0.5\\\\\theta=30^{\circ}$

So, the angle between A and B is 30 degrees.

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