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

Question

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

Answer 1

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° = \(\frac{\sqrt3}2\))