First slide

Scalars and Vectors

Question

The angle between $\bar{A} = 3 \hat{i} + 4 \hat{j} & \bar{B} = \hat{i} + \hat{j}$ in degrees NEARLY is _______

Easy

Solution

$C O S θ = \frac{\vec{A} . \vec{B}}{|A| |B|} = \frac{3 \times 1 + 4 \times 1}{5 \times \sqrt{2}} = 1 θ = 45^{0}$

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