Question

Two vectors have magnitudes 2 m and 3 m. The angle between them is ${60}^0$. Find

(a) the scalar product of the two vectors,

(b) the magnitude of their vector product.

Answer 1

$\overrightarrow{|a|}=2m;\overrightarrow{|a|}=3m$

Angle between them,

(a) $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{|a|}.\overrightarrow{|b|}cos{60}^0$

= $2\times 3\times \frac{1}{2}=3m^2$

(b) $|\overrightarrow{a}\times \overrightarrow{b}|=\overrightarrow{|a|}\overrightarrow{|b|}sin{60}^0$

= $2\times 3\times \frac{\sqrt{3}}{2}$

= $3\sqrt{3}{m}^2$