Question

24. Find the angle between the vector Ā = -1 + 2
j - k and B = i + j - 2K *
(1 Point)
120°
25°
30°
60°​

Answer 1

Answer:

(D)

Explanation:

Using dot product:

$\\\tt a.b = |a||b| \cos(\theta)\\(-i + 2j -k).(i+j-2k) = \sqrt{1^{2} + 2^{2} + 1^{2} } \sqrt{1^{2} + 1^{2} + 2^{2} } \cos(\theta)\\(-1+2+2) = 6 \cos(\theta)\\3 = 6 \cos(\theta)\\\cos(\theta) =\dfrac{1}{2} \\\theta = 60^{\circ}$