Question

If a unit vector $\overline{a}$ makes angles $\frac{\pi }{3}$ with $\hat{i}$ , $\frac{\pi }{4}$ with $\hat{j}$ and an acute angle $\theta$ with $\hat{k}$, find $\theta$ and hence, the component of $\overrightarrow{a}$.

Answer 1

Let us take a unit vector
$\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k}$
So magnitude of $\overrightarrow{a}=\left|\overrightarrow{a}\right|=1$
Angle of $\overrightarrow{a}$ with $\hat{i}=\frac{\pi }{3}$
$\overrightarrow{a}\hat{i}=\left|\overrightarrow{a}\right|\left|\overrightarrow{i}\right|\cos \frac{\pi }{3}$
$\Rightarrow (x\hat{i}+y\hat{j}+z\hat{k}).\hat{i}=1\times 1\times \frac{1}{2}$
$\Rightarrow (x\hat{i}+y\hat{j}+z\hat{k}).(1\hat{i}+0\hat{j}+0\hat{k})=\frac{1}{2}$
$\Rightarrow (x\times 1)+(y\times 0)+(z\times 0)=\frac{1}{2}$
$\Rightarrow x+0+0=\frac{1}{2}$
$\Rightarrow x=\frac{1}{2}$
Angle of $\overrightarrow{a}$with $\hat{j}=\frac{\pi }{4}$
$\overrightarrow{a}\hat{j}=\left|\overrightarrow{a}\right|\left|\overrightarrow{j}\right|\cos \frac{\pi }{4}$
$\Rightarrow (x\hat{i}+y\hat{j}+z\hat{k}).\hat{j}=1\times 1\times \frac{1}{\sqrt{2}}$
$\Rightarrow (x\hat{i}+y\hat{j}+z\hat{k}).(0\hat{i}+1\hat{j}+0\hat{k})=\frac{1}{\sqrt{2}}$
$\Rightarrow (x\times 0)+(y\times 1)+(z\times 0)=\frac{1}{\sqrt{2}}$
$\Rightarrow 0+y+0=\frac{1}{\sqrt{2}}$
$\Rightarrow y=\frac{1}{\sqrt{2}}$
Angle of $\overrightarrow{a}$ with $\hat{k}=\theta$
$\overrightarrow{a}.\hat{k}=\left|\overrightarrow{a}\right|\left|\hat{k}\right|\times \cos \theta$
$(x\hat{i}+y\hat{j}+z\hat{k}).(0\hat{i}+0\hat{j}+1\hat{k})=1\times 1\times \cos \theta$
$\Rightarrow (x\times 0)+(y\times 0)+(z\times 1)=\cos \theta$
$\Rightarrow 0+0+z=\cos \theta$
$\Rightarrow z=\cos \theta$
Now, magnitude of $\overrightarrow{a}=\sqrt{x^2+y^2+z^2}$
$\Rightarrow 1=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2+{\cos }^2\theta }$ (As $|\overrightarrow{a}=1|$)
$\Rightarrow 1=\sqrt{\frac{1}{4}+\frac{1}{2}+{\cos }^2\theta }$
$\Rightarrow 1=\sqrt{\frac{3}{4}+{\cos }^2\theta }$
$\Rightarrow {\left(\sqrt{\frac{3}{4}+{\cos }^2\theta }\right)}^2=1^2$
$\Rightarrow \frac{3}{4}+{\cos }^2\theta =1$
$\Rightarrow {\cos }^2\theta =\frac{1}{4}$
$\Rightarrow \cos \theta =\pm \sqrt{\frac{1}{4}}$
$\Rightarrow \cos \theta =\pm \frac{1}{2}$
Since $\theta$ is given an acute angle
So, $\theta <90°$
$\therefore \theta$ is in first quadrant
& $\cos \theta$ is positive in first quadrant
So, $\cos \theta =+\frac{1}{2}$
$\Rightarrow \theta =60°=\frac{\pi }{3}$
Also, $z=\cos \theta =\cos 60°=\frac{1}{2}$
Hence, $x=\frac{1}{2},y=\frac{1}{\sqrt{2}}$ & $z=\frac{1}{2}$
The required vector $\overrightarrow{a}$ is $\frac{1}{2}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k}$
So, components of $\overrightarrow{a}$ are $\frac{1}{2},\frac{1}{\sqrt{2}}$ &$\frac{1}{2}$