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 $a=2 \mathrm{~m}, b=3 \mathrm{~m}$.
$\theta=60^{\circ}$ 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 $=a \times b=$ $2 \times 3 \times \sin \left(60^{\circ}\right)=3 \sqrt{3} \mathrm{~m}^{2}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.