Question

A position vector of a point is (2î +2j)m
(a) Find magnitude of this vector
(b) Find angle with x-axis
(c) Find the volume of cone which is generated when a line segment representing this position vector
is rotated about x-axis with one end remaining fixed at origin.​

Answer 1

Given that,

A position vector of point is

$\vec{r}=(2i+2j)\ m$

(a). We need to calculate the magnitude of this vector

Using formula of magnitude of the vector

$r=\sqrt{2^2+2^2}$

$r=\sqrt{8}$

$r=2\sqrt{2}\ m$

(b). We need to calculate the angle with x-axis

Using formula of direction

$\theta=\tan^{-1}\dfrac{\text{component of j}}{\text{component of i}}$

Put the value in to the formula

$\theta=\tan^{-1}\dfrac{2}{2}$

$\theta=45^{\circ}$

(c). When a line segment representing this position vector  is rotated about x-axis with one end remaining fixed at origin

The radius of the cone formed by the position vector is 2 unit.

The height of the cone = 2 unit

We need to calculate the volume of the cone

Using formula of volume of cone

$V=\dfrac{1}{3}\pi r^2 h$

Put the value into the formula

$V=\dfrac{1}{3}\times\pi\times2^2\times2$

$V=8.37\ cubic\ unit$

Hence, (a). The magnitude of this vector is 2√2 unit.

(b). The angle with x-axis is 45°

(c). The volume of cone is 8.37 cubic unit.