Question

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

Answer 1

Hint
In this problem it is given that the bird moves in some angle and some angle. Here it is given that the angle with respect to two axes. By using the direction of cosine formula, the third can be determined and then the velocity vector is represented in rectangular form.

Complete step by step solution
The given two angles are assumed to be in x-axis and y-axis, so the two angles are taken as $\alpha $ and $\beta $.
Given that The velocity, $V = 20\,m{s^{ - 1}}$ , $\alpha = {60^ \circ }$ and $\beta = {60^ \circ }$
By using the direction of cosine formula,
${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1\,.................\left( 1 \right)$
Substitute the $\alpha $ and $\beta $ value in the equation (1), then the above equation is written as,
${\cos ^2}\left( {{{60}^ \circ }} \right) + {\cos ^2}\left( {{{60}^ \circ }} \right) + {\cos ^2}\gamma = 1$
The above equation can also be written as,
${\left( {\cos \left( {{{60}^ \circ }} \right)} \right)^2} + {\left( {\cos \left( {{{60}^ \circ }} \right)} \right)^2} + {\cos ^2}\gamma = 1$
By trigonometry, the value of $\cos {60^ \circ } = \dfrac{1}{2}$, and substitute this value in the above equation, then,
${\left( {\dfrac{1}{2}} \right)^2} + {\left( {\dfrac{1}{2}} \right)^2} + {\cos ^2}\gamma = 1$
By squaring the terms,
$\left( {\dfrac{1}{4}} \right) + \left( {\dfrac{1}{4}} \right) + {\cos ^2}\gamma = 1$
By adding the terms, then the above is written as,
$\dfrac{2}{4} + {\cos ^2}\gamma = 1$
And by cancelling the terms, then the above equation is written as,
$\dfrac{1}{2} + {\cos ^2}\gamma = 1$
By keeping the ${\cos ^2}\gamma $ in one side and other terms in another side, then,
${\cos ^2}\gamma = 1 - \dfrac{1}{2}$
Then the above equation is written as,
${\cos ^2}\gamma = \dfrac{1}{2}$
On further steps,
$\cos \gamma = \dfrac{1}{{\sqrt 2 }}$
The velocity vector for rectangular form is,
$\vec V = V\cos \alpha \hat i + V\cos \beta \hat j + V\cos \gamma \hat k\,.............\left( 2 \right)$
Substituting the values of $\cos \alpha $, $\cos \beta $, $\cos \gamma $ and velocity $V$ in the equation (2), then
$\vec V = 20 \times \dfrac{1}{2}\hat i + 20 \times \dfrac{1}{2}\hat j + 20 \times \dfrac{1}{{\sqrt 2 }}\hat k\,$
By cancelling the terms in above equation,
$\vec V = 10\hat i + 10\hat j + 20 \times \dfrac{1}{{\sqrt 2 }}\hat k\,$
In the above equation the value $20$ is written as $\left( {10 \times 2} \right)$ for further calculation, then
$\vec V = 10\hat i + 10\hat j + \left( {10 \times 2} \right) \times \dfrac{1}{{\sqrt 2 }}\hat k\,$
Now the term $2$ is written as $\left( {\sqrt 2 \times \sqrt 2 } \right)$ for further calculation, then the above equation is written as,
$\vec V = 10\hat i + 10\hat j + \left( {10 \times \sqrt 2 \times \sqrt 2 } \right) \times \dfrac{1}{{\sqrt 2 }}\hat k\,$
By cancelling the same terms in numerator and denominator, then the above equation is written as,
$\vec V = 10\hat i + 10\hat j + 10\sqrt 2 \hat k\,$
Therefore, the rectangular vector form of the velocity is $\vec V = 10\hat i + 10\hat j + 10\sqrt 2 \hat k\,$.

Note
While solving this problem we have to give more concentration in converting the term from ${\cos ^2}{60^ \circ }$ to ${\left( {\cos {{60}^ \circ }} \right)^2}$, in trigonometry this conversion of power is very important. And we have to give more focus on cancelling the terms, here to cancel the term $\sqrt 2 $ in denominator, some value is written in another form.