Question

If a unit vector $\vec{a}$ makes an angles $\frac{\pi}{3}$ with $\hat{i}$,$\frac{\pi}{4}$ with $\hat{j}$ and an acute angle $θ$ with $\hat{k}$ then find $θ$ and hence,the compounds of $\vec{a}$.

Answer 1

Let unit vector $\vec{a}$ have $(a_{1},a_{2},a_{3})$ components.
$\vec{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}$
Since $\vec{a}$ is a unit vector,|$\vec{a}$|$=1$.
Also, it is given that $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}$,$\frac{\pi}{4}$ with $\hat{j}$,and an acute angle $θ$ with $\hat{k}$.
Then, we have:
$cos\frac{\pi}{3}=\frac{a_{1}}{|\vec{a}|}$
$⇒\frac{1}{2}=a_{1} [|\vec{a}|=1]$
$cos\frac{\pi}{4}=\frac{a_{2}}{|\vec{a}|}$
$⇒\frac{1}{\sqrt{2}}=a_{2} \space\space\space\space\space[|\vec{a}|=1]$
Also,$cosθ=\frac{a_{3}}{|\vec{a}|}.$
$⇒a_{3}=cosθ$
Now,
$|a|=1$
$⇒\sqrt{a^{2}_{1}+a^{2}_{2}+a^{2}_{3}}=1$
$⇒(\frac{1}{2})^{2}+(\frac{1}{\sqrt{2}})^{2}+cos^{2}θ=1$
$⇒\frac{1}{4}+\frac{1}{2}+cos^{2}cosθ=1$
$⇒\frac{3}{4}+cos^{2}θ=1$
$⇒cos^{2}θ=1-\frac{3}{4}=\frac{1}{4}$
$⇒cosθ=\frac{1}{2}⇒θ=\frac{\pi}{3}$
$∴a_{3}=cos\frac{\pi}{3}=\frac{1}{2}$
Hence,$θ=\frac{\pi}{3}$ and the components of $\vec{a}$ are($(\frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2})$).