Question

vector P=i+2j+3k and Q vector=2i-j-k. Find the angle between 2P+Q and P-2Q vectors.

Answer 1

Sol : 2P+Q = 2(i+2j+3k) + 2i-j-k = 4i+3j+5k
and P-2Q = i+2j+3k - (4i-2j-2k) = -3i+4j+5k

Now angle b/w them is given by using : cos(theta) = a.b/|a|×|b| ,where a and b are two vectors and theta is the angle b/w them.

So,
cos(theta) = [(4i+3j+5k).(-3i+4j+5k)]/(√50×√50)
where √50 is magnitude of both required vectors.

So, cos(theta) = 25/50 = 1/2
or theta = cos^-1(1/2) = π/3 or 60°
So angle b/w required vectors is 60° or π/3

Answer 2

Explanation:

angle between the required vector is 60°