Question

Find a vector $\overrightarrow{a}$ of magnitude $5\sqrt{2}$ making an angle of $\dfrac{\pi }{4}$ with x – axis, $\dfrac{\pi }{2}$ with the y – axis and an acute angle $\theta$ with the z – axis.

Answer 1

Hint: To find the vector, we need to use the concept of direction cosines of a vector. Thus, we will define the property of directional cosines. Then we will substitute the given angles into that property and find the angle $\theta$. Then we will find the vector $\overrightarrow{a}$ which is given as $\overrightarrow{a}=\left| \overrightarrow{a} \right|\left( l\hat{i}+m\hat{j}+n\hat{k} \right)$, where l, m and n are the direction cosines of the vector, $\left| \overrightarrow{a} \right|$ is the magnitude of vector $\overrightarrow{a}$ and $\hat{i}$, $\hat{j}$ and $\hat{k}$ are unit vectors in the direction x, y and z axes respectively.

Complete step-by-step answer:

The direction cosines (or directional cosines) of a vector are the cosines of the angles between the vector and the three coordinate axes, i.e. x, y and z axes.

Let p, q and r be the angle made by a vector with the three coordinates axes x, y and z respectively.

Thus, l = cos p, m = cos q and n = cos r, where l, m and n are the directional cosines of the vector.

For l, m and n to be directional cosines, they must follow one condition given as ${{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1$.

We are given that $\overrightarrow{a}$ makes an angle of $\dfrac{\pi }{4}$ with x – axis, $\dfrac{\pi }{2}$ with the y – axis and an acute angle $\theta$ with the z – axis.

Thus, l = $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$, m = $\cos \dfrac{\pi }{2}=0$ and n = cos$\theta$.

\Rightarrow {( \dfrac{1}{\sqrt{2}})}^{2}}+0+{{\cos }^{2}}\theta =1

$\Rightarrow \cos \theta =\dfrac{1}{\sqrt{2}}$

Hence, the vector makes an angle of $\dfrac{\pi }{4}$ with z – axis and n = $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$.

The vector $\overrightarrow{a}$ is given as $\overrightarrow{a}=\left| \overrightarrow{a} \right|\left( l\hat{i}+m\hat{j}+n\hat{k} \right)$, where l, m and n are the direction cosines of the vector, $\left| \overrightarrow{a} \right|$ is the magnitude of vector $\overrightarrow{a}$ and $\hat{i}$, $\hat{j}$ and $\hat{k}$ are unit vectors in the direction x, y and z axes respectively.

We are given that $\left| \overrightarrow{a} \right|=5\sqrt{2}$

\Rightarrow \overrightarrow{a}=5\sqrt{2}( \dfrac{1}{\sqrt{2}}\hat{i}+ ( 0 \right)\hat{j}+\dfrac{1}{\sqrt{2}}\hat{k})

$\Rightarrow \overrightarrow{a}=5\left( \hat{i}+\hat{k} \right)$

Note: The direction cosines (or directional cosines) of a vector are the cosines of the angles between the vector and the three coordinate axes, i.e. x, y and z axes. Let p, q and r be the angle made by a vector with the three coordinates axes x, y and z respectively. Thus, l = cos p, m = cos q and n = cos r, where l, m and n are the directional cosines of the vector.