Question

A vector having magnitude 16 units is rotated through an angle 60° about its tail. The magnitude of change in vector is ​


Help.

Answer 1

Answer:

hey simmi ...I am also in class11th ...u can discuss with me on any topic ....

mark me brainliest if it helps u

Answer 2

Answer:

The magnitude of change in vector is 16 unit.

Explanation:

Given that,

Magnitude of vector a= 16 unit

Angle $\theta= 60^{\circ}$

We need to calculate the magnitude of change in vector

From the triangle law of vectors

$\vec{PQ}=\vec{OQ}-\vec{OP}$

$\vec{PQ}=\vec{OQ}+(-\vec{OP})$

Angle between $\vec{OQ}$ and $\vec{-OP}=180^{\circ}-\theta$

$(\Delta a)^2=(a)^2+(a)^2+2\times(a)^2\cos(180^{\circ}-\theta$

Put the value into the formula

$(\Delta a)^2=(16)^2+(16)^2+2\times(16)^2\times\cos(180^{\circ}-\theta$

$(\Delta a)^2=2\times16^2+2\times16^2\times\cos\theta$

$(\Delta a)^2=2\times16^2(1-\cos\theta)$

$(\Delta a)^2=2\times16^2\times(1-1+\dfrac{2\sin^2\theta}{2})$

$(\Delta a)^2=4\times16^2\times\dfrac{\sin^2\theta}{2}$

$(\Delta a)^2=4\times16^2\times\dfrac{\sin^2\theta}{2}$

Put the value of θ

$(\Delta a)^2=4\times16^2\times\dfrac{\sin^2(60)}{2}$

$\Delta a=2\times16\times\dfrac{\sin(60)}{2}$

$\Delta a=2\times16\times\sin30$

$\Delta a=16\ unit$

Hence, The magnitude of change in vector is 16 unit.