Question

The X-component of $\overrightarrow{a}$ is twice of its Y-component. If the magnitude of the vector is $5\sqrt{2}$ and it makes an angle of $135{}^\circ $ with z-axis then the components of vector is
A.$2\sqrt{3},\sqrt{3},-3$
B.$2\sqrt{6},\sqrt{6},-6$
C.$2\sqrt{5},\sqrt{5},-5$
D.$2\sqrt{2},\sqrt{2},-2$

Answer 1

Hint: As a first step, you could find the z-component of the given vector from the given angle made by the vector with the z-axis. Now you could recall the expression for the magnitude of the vector in terms of its x, y, and z components. Then you could substitute accordingly using the given relation between the x and y components and thus find the answer.

Formula used:
The magnitude of a vector,
$\left| \overrightarrow{a} \right|=\sqrt{{{a}_{x}}^{2}+{{a}_{y}}^{2}+{{a}_{z}}^{2}}$

Complete step by step solution:
In the question, we are given that x-component of $\overrightarrow{a}$ vector is twice the y-component, that is,
$\overrightarrow{{{a}_{x}}}=2\overrightarrow{{{a}_{y}}}$ ………………………………………….. (1)
The magnitude of the vector $\overrightarrow{a}$ is given as, $5\sqrt{2}$ and the angle made by the vector with the z-axis is given as $135{}^\circ $. We know that the z-component of the vector is given by,
${{a}_{z}}=a\cos \gamma $
Where, $a$ is the magnitude of the vector and $\gamma $ is the angle made by the vector with the z-axis. So,
${{a}_{z}}=5\sqrt{2}\cos 135{}^\circ $
$\Rightarrow {{a}_{z}}=5\sqrt{2}\times \dfrac{-1}{\sqrt{2}}$
$\therefore {{a}_{z}}=-5$ ………………………………… (2)
We know that vector $\overrightarrow{a}$ is given by,
$\overrightarrow{a}={{a}_{x}}\widehat{i}+{{a}_{y}}\widehat{j}+{{a}_{z}}\widehat{k}$
Magnitude of this vector is given by,
$\left| \overrightarrow{a} \right|=\sqrt{{{a}_{x}}^{2}+{{a}_{y}}^{2}+{{a}_{z}}^{2}}$
Substituting the magnitude of vector $\overrightarrow{a}$ and that of its z-component along with the relation (1), we get,
$5\sqrt{2}=\sqrt{{{\left( 2{{a}_{y}} \right)}^{2}}+{{a}_{y}}^{2}+{{\left( -5 \right)}^{2}}}$
$\Rightarrow {{\left( 5\sqrt{2} \right)}^{2}}={{\left( 2{{a}_{y}} \right)}^{2}}+{{a}_{y}}^{2}+25$
$\Rightarrow 50=5{{a}_{y}}^{2}+25$
$\Rightarrow {{a}_{y}}^{2}=5$
$\therefore {{a}_{y}}=\sqrt{5}$
From (1) we get the x-component of the given vector,
${{a}_{x}}=2{{a}_{y}}=2\sqrt{5}$
Therefore, we found the x, y and z components of the given vectors as $2\sqrt{5},\sqrt{5},-5$ respectively. Hence, option C will be the right answer.

Note: The x and y components can be also expressed in terms of the angle made by the vector with the x and y axis respectively as,
${{a}_{x}}=a\cos \alpha $
${{a}_{y}}=a\cos \beta $
Where, $\alpha $ and $\beta $ are the angles made by the vector with x and y axis respectively. This problem could be solved alternatively by using the relation,
${{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1$