Question

If a straight line makes an angle ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$ with each of the positive x, y and z-axis, a vector parallel to that line is?

• A. $\overline{i}$
• B. $\overline{i}+\overline{j}$
• C. $\hat{j}+\overline{k}$
• D. $\overline{i}+\overline{j}+\overline{k}$

Answer 1

The correct option is D $\overline{i}+\overline{j}+\overline{k}$

Let the vector be $x\hat{i}+y\hat{j}+z\hat{k}$

makes ${\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$ with $x,y$ and $z$ axis

$\cos \theta =(x\hat{i}+y\hat{j}+z\hat{k}).\hat{i}$

$\cos \left({\cos }^{-1}\frac{1}{\sqrt{3}}\right)=x$

$x=\frac{1}{\sqrt{3}}$

$\cos \theta =(x\hat{i}+y\hat{j}+z\hat{k}).\hat{j}$

$\cos \left({\cos }^{-1}\frac{1}{\sqrt{3}}\right)=y\Rightarrow y=\frac{1}{\sqrt{3}}$

$\cos \theta (x\hat{i}+y\hat{j}+z\hat{k})$

$\cos \left({\cos }^{-1}\frac{1}{\sqrt{3}}\right)=z\Rightarrow z=\frac{1}{\sqrt{3}}$

$\therefore$ the vector is $\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$

Which is parallel to $\hat{i}+\hat{j}+\hat{k}$