v = −i+2j+k, w = −3i+6j+3k Find the angle θ between the vectors v and w

Question

v = −i+2j+k, w = −3i+6j+3k Find the angle θ between the vectors v and w

Answer 1

To find the angle between two vectors, use:

 

a•b = ||a|| ||b|| cosθ

 

v = <-1, 2, 1>

w = <-3, 6, 3>

 

v•w = -1(-3) + 2(6) + 1(3) = 3+12+3 = 18

 

||v|| = √((-1)² + 2² + 1²) = √(1+4+1) = √6

 

||w|| = √((-3)^{2 +} 6² + 3²) = √(9+36+9) = √54 = 3√6

 

cos θ = (v•w) / (||v|| ||w||) = 18/( √6 * 3√6) = 18/18 = 1

 

This simplifies to cos θ = 1

 

θ = cos^-1(1) = 0

 

Therefore, the angle between the 2 vectors is 0.  This makes sense because the vectors are scalar multiples of each other:  v = 3w

 

 

Hope this helps!

Cara Marie