First slide

Vectors

Question

The angle between the vectors 2i+3j+k and -3i+6k is

Easy

A

0°
B

45°
C

60°
D

90°

Solution

$\cos θ = \frac{A \cdot B}{AB}$
A.B = (2 i + 3j - k) . (- 3i + 6 k)
=2(-3)+3(0) +1(6) = 0
$∴ \cos θ = \frac{0}{\sqrt{14} \times \sqrt{45}} = 0 or θ = 90^{\circ}$

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