First slide

Scalars and Vectors

Question

The angle between vectors $\vec{A} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$ and $\vec{B} = 9 \hat{i} + 12 \hat{j} + 15 \hat{k}$ is :

Difficult

A

Zero
B

$45^{0}$
C

$90^{0}$
D

$180^{0}$

Solution

$\cos θ = \frac{\vec{A} . \vec{B}}{A B}$
$= \frac{3 \times 9 + 4 \times 12 + 5 \times 15}{\sqrt{{(3)}^{2} + {(4)}^{2} + {(5)}^{2}} \times \sqrt{{(9)}^{2} + {(12)}^{2} + {(15)}^{2}}}$
$= \frac{150}{\sqrt{50} \times \sqrt{450}}$
$= \frac{150}{150} = 1$
$θ = \cos^{- 1} 1 = 0^{0}$

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