# Vectors a and b are inclined at an angle = 120. If |a| = |b| = 2 , then find value of [( a +3b )*(3a + b )]^ 2

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Vectors a and b are inclined at an angle = 120. If |a| = |b| = 2 , then find value of [( a +3b ) x (3a + b )]2

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Angle between a and b is θ  = 120° and |a| = |b| = 2

Now, [(a + 3b) x (3a + b)]2 = [a x 3a + a x b + 9(b x a) + 3b x b]2

= [a x b - 9(a x b)]2 ($\because$ a x a = 0, b x b = 0 and b x  a = -(a x b))

= [-8(a x b)]2 = 64(a x b)2

= 64 (|a|.|b| sin θ $\hat n$)2 ($\because$ a x b = |a|.|b| sin θ $\hat n$ )

= 64 |a|2.|b|2 sin2θ ($\hat n^2=\hat n.\hat n=1$

= 64 x 22 x 22 x sin2120° ($\because$ |a| = |b| = 2)

= 64 x 16 x 3/4 = 64 x 12 = 768 ($\because$ sin120° = sin 60° = $\frac{\sqrt3}2$)