Question:
Two vectors have magnitudes $2 \mathrm{~m}$ and $3 \mathrm{~m}$. The angle between them is $60^{\circ}$. Find (a) the scalar product of the two vectors, (b) the magnitude of their vector product?
Solution:
We have $a=2 m, b=3 m$.
$\Theta=60^{\circ}$ is the angle between the two vectors
Scalar product between the two vectors $=a \cdot b=$
$2 \times 3 \times \cos \left(60^{\circ}\right)=3 \mathrm{~m}^{2}$
Vector product between the two vectors $=a \times b=$
$2 \times 3 \times \sin \left(60^{\circ}\right)=3 \sqrt{3} \mathrm{~m}^{2}$