Question

If vectors$a_1=x\hat{i}-\hat{j}+\hat{k}$ and $a_2=\hat{i}+y\hat{j}+z\hat{k}$ are collinear, then a possible unit vector parallel to the vector$x\hat{i}+y\hat{j}+z\hat{k}$ is

• A. $\left(\frac{1}{\sqrt{2}}\right)\left(-\hat{j}+\hat{k}\right)$
• B. $\left(\frac{1}{\sqrt{2}}\right)\left(\hat{i}–\hat{j}\right)$
• C. $\left(\frac{1}{\sqrt{3}}\right)\left(\hat{i}-\hat{j}+\hat{k}\right)$
• D. $\left(\frac{1}{\sqrt{3}}\right)\left(\hat{i}+\hat{j}–\hat{k}\right)$

Answer 1

The correct option is C

$\left(\frac{1}{\sqrt{3}}\right)\left(\hat{i}-\hat{j}+\hat{k}\right)$

Explanation for correct option:

Finding the value of the expression:

Vectors$a_1=x\hat{i}-\hat{j}+\hat{k}$ and $a_2=\hat{i}+y\hat{j}+z\hat{k}$ are collinear

$\frac{x}{1}=\frac{-1}{y}=\frac{1}{z}=\lambda \left(let\right)$

According to the question, a unit vector is parallel to$x\hat{i}+y\hat{j}+z\hat{k}$,

$x\hat{i}+y\hat{j}+z\hat{k}$ $=\frac{\pm \left[\lambda \hat{i}–\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.}\right)\hat{j}+\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.}\right)\hat{k}\right]}{\sqrt{{\lambda }^2+\left(2/{\lambda }^2\right)}}$

Now, for $\lambda =1$ it is,

$x\hat{i}+y\hat{j}+z\hat{k}=\frac{\pm \left(\hat{i}–\hat{j}+\hat{k}\right)}{\sqrt{3}}$

Hence, the correct option is C.