Question

A bird moves with a velocity of $20m{s}^{-1}$ in a direction making an angle of $60°$ with the eastern line and $60°$ with vertical upward. Represent the velocity vector in rectangular form.

Answer 1

Step 1: Given Data

The magnitude of the velocity $\left|v\right|=20m/s$

Eastern angle of the velocity $\alpha =60°$

The vertical angle of the velocity $\beta =60°$

Let the horizontal angle be $r$.

Step 2: Calculate the horizontal angle

We know that,

${\cos }^2\alpha +{\cos }^2\beta +{\cos }^{2}r=1$

$\Rightarrow {\cos }^{2}60°+{\cos }^{2}60°+{\cos }^{2}r=1$

$\Rightarrow {\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2+{\cos }^{2}r=1$

$\Rightarrow {\cos }^{2}r=1-\frac{1}{2}$

$\Rightarrow \cos r=\pm \frac{1}{\sqrt{2}}$

Step 3: Estimate the Vector

The velocity vector can be given as,

$\overrightarrow{v}=v\cos \alpha \hat{i}+v\cos \beta \hat{j}+v\cos r\hat{k}$

$\Rightarrow \overrightarrow{v}=20\times \frac{1}{2}\hat{i}+20\times \frac{1}{2}\hat{j}+20\times \frac{1}{\sqrt{2}}\hat{k}$

$\Rightarrow \overrightarrow{v}=10\hat{i}+10\hat{j}+10\sqrt{2}\hat{k}$

Hence, the velocity vector in rectangular form is $10\hat{i}+10\hat{j}+10\sqrt{2}\hat{k}$.