First slide

Scalars and Vectors

Question

The angle between the vectors $(\hat{i} + \hat{j})$ and $(\hat{j} + \hat{k})$ is:

Moderate

A

$90^{0}$
B

$60^{0}$
C

$30^{0}$
D

$45^{0}$

Solution

Let, $\vec{A} = \hat{i} + \hat{j}$ and $\vec{B} = \hat{j} + \hat{k}$
$\vec{A} . \vec{B} = (\hat{i} + \hat{j}) . (\hat{j} + \hat{k}) = 1$
Also, $| \vec{A} | = \sqrt{{(1)}^{2} + {(1)}^{2}} = \sqrt{2}$
$| \vec{B} | = \sqrt{{(1)}^{2} + {(1)}^{2}} = \sqrt{2}$
Now $\cos θ = \frac{\vec{A} . \vec{B}}{| \vec{A} | . | \vec{B} |} = \frac{1}{\sqrt{2} \times \sqrt{2}} = \frac{1}{2}$
$θ = \cos^{- 1} (\frac{1}{2}) = 60^{0}$

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