Two vectors have magnitudes 2 m and 3m. The angle between them is 60°. Find (a) the scalar product of the two vectors, (b) the magnitude of their vector product.

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Solution

Let the two vectors be $|\vec{a}| = 2 m and |\vec{b}| = 3 m$ .
Angle between the vectors, θ = 60°

(a) The scalar product of two vectors is given by $\vec{a} . \vec{b} = |\vec{a}| . |\vec{b}| \cos θ °$ .

∴ $\vec{a} . \vec{b} = |\vec{a}| . |\vec{b}| \cos 60 °$
$= 2 \times 3 \times \frac{1}{2} = 3 m^{2}$

(b) The vector product of two vectors is given by $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{a}| \sin θ °$ .
∴ $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{a}| \sin 60 °$
$= 2 \times 3 \times \frac{\sqrt{3}}{2} = 3 \sqrt{3} m^{2}$

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