Two vectors , vector A = î +4ĵ- 3k^{^}& B=3î-ĵ-2k^{^}lie in the plane of this paper. Find the cross product of A & B. Hence determine the angle bt. A&B

Cross product is written as

C = A X B = i [A_yB_z - A_zB_y] + j [A_zB_x - A_xB_z] + k [A_xB_y - A_yB_x]

and

A = i + 4j - 3k

B = 3i - j - 2k

so,

C = i [{4 X (-2)} - {(-3) X (-1)}] + j [{3 X (-3)} - {1 X (-2)}] + k[{1 X (-1)} - {4 X 3}]

or

C = i [-8 -3] + j [-9 + 2] + k [-1 -12]

so,

the cross product of A and B is

C = -11i - 7j - 13k

the angle between the vectors can be given as

A . B = |A| |B| Cosθ

or COsθ = |A| |B| / [A . B]

where |A| = [1 + 16 + 9]^1/2

|B| = [9 + 1 + 4]^1/2

and

A . B = [i + 4j - 3k].[3i - j - 2k]

thus,

Cosθ =  [1 + 16 + 9]^1/2 X  [9 + 1 + 4]^1/2 /  [i + 4j - 3k].[3i - j - 2k]

which can be calculated...