Question

Find the sine of the angle between the vectors i+3j+2k and 2i-4j+k

Answer 1

Answer:

Explanation:

a =3i +j+2k

b=2i-2j+4k

axb = i(4+4)-j (12-4)+k (-6 -2)

      =8i -8j-8k

|axb| = (64+64+64)1/2

        = (192)1/2

|a| = (9+1+4)1/2 =141/2

|b|= (4+4+16)1/2 =241/2

SinQ =|axb|/|a| |b|

        = (192)1/2 /(14)1/2 (24)1/2