Question

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.

Answer 1

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

(a) The scalar product of two vectors is given by $\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|\cos \theta °$.

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

(b) The vector product of two vectors is given by $\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\left|\overrightarrow{a}\right|\left|\overrightarrow{a}\right|\sin \theta °$.
∴ $\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\left|\overrightarrow{a}\right|\left|\overrightarrow{a}\right|\sin 60°$
$$\begin{gathered} =2\times 3\times \frac{\sqrt{3}}{2} \\ =3\sqrt{3}{\mathrm{m}}^2 \end{gathered}$$