Find the magnitude of vector (a) -4.00i-1.00j+2.00k (b) 2.00i-6.00j+3.00k (c) angle between these two vectors?

Answer 1

Define the two vectors

a = -4i - j + 2k
b = 2i - 6j + 3k

|| a || = √[(-4)² + (-1)² + 2²] = √(16 + 1 + 4) = √21
|| b || = √[2² + (-6)² + 3²] = √(4 + 36 + 9) = √49 = 7

The angle θ between vectors a and b can be determined from the definition of the dot product. First take the dot product of a and b.

a • b = <-4, -1, 2> • <2, -6, 3> = (-4)*2 + (-1)(-6) + 2*3
a • b = -8 + 6 + 6 = 4

The dot product can also be given by:

a • b = || a || || b || cosθ

cosθ = (a • b) / (|| a || || b ||) = 4 / [(√21)*7]
cosθ = 4 / (7√21)

θ = arccos[4 / (7√21)] ≈ 82.836803°