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.

#### Solution

Let the two vectors be \[\left| \vec{a} \right| = 2 m\text { and } \left| \vec{b} \right| = 3 m\].

Angle between the vectors, θ = 60°

(a) The scalar product of two vectors is given by \[\vec{a} . \vec{b} = \left| \vec{a} \right| . \left| \vec{b} \right| \cos\theta^\circ\]

∴ \[\vec{a} . \vec{b} = \left| \vec{a} \right| . \left| \vec{b} \right| \cos 60^\circ\]

\[= 2 \times 3 \times \frac{1}{2} = 3 m^2\]

(b) The vector product of two vectors is given by \[\left| \vec{a} \times \vec{b} \right| = \left| \vec{a} \right| \left| \vec{a} \right| \sin\theta^\circ\].

\[= 2 \times 3 \times \frac{\sqrt{3}}{2}\]

\[ = 3\sqrt{3} m^2\]