Question

sum of three vectors is zero such that two vectors out of them have equal magnitude and magnitude of third vector is equal to √2 times that of first vector.....find the angle between them......

Answer 1

As sum of three vectors is zero,they form a triangle.
Let The resultant is the third vector and magnitude of two equal vectors is a.

so magnitude of third vector is a√2.
Let angle between two equal vectors is alpha.

Now we have......
$a \sqrt{2} = \sqrt{a {}^{2} + a {}^{2} + 2a.acos \alpha }$
$a \sqrt{2} = \sqrt{2a {}^{2} + 2a {}^{2}cos \alpha }$
a√2=√2a√(1+cos alpha).
=> √(1+cos alpha)=1

on squaring....

1+cos alpha=1
=>cos alpha=0
=> alpha=90°

so angle between two equal vectors is 90°.

Now we can find angle between vectors by above image.