Question

The unit vector in the direction of the vector $a=\hat{i}+\hat{j}+2\hat{k}$ is

• A. $\frac{1}{\sqrt{6}}\hat{i}+\frac{1}{\sqrt{6}}\hat{j}+\frac{1}{\sqrt{6}}\hat{k}$
• B. $\frac{2}{\sqrt{6}}\hat{i}+\frac{3}{\sqrt{6}}\hat{j}+\frac{1}{\sqrt{6}}\hat{k}$
• C. $\frac{1}{\sqrt{6}}\hat{i}+\frac{1}{\sqrt{6}}\hat{j}+\frac{2}{\sqrt{6}}\hat{k}$
• D. None of these

Answer 1

Answer: (C)

Unit vector in the direction of given vector can be calculated by formula $\hat{a}=\frac{a}{|a|}$ where, $|a|=$ magnitude of given vector a.
Given, $a=\hat{i}+\hat{j}+2\hat{k}$ comparing with $x=a\hat{i}+b\hat{j}+c\hat{k}$, we get $a=1,b=1,c=2$
Magnitude of vector $a=|a|=\sqrt{a^2+b^2+c^2}$
$=\sqrt{1^2+1^2+2^2}=\sqrt{6}$
$\therefore$ The unit vector $\hat{a}$ in the direction of vector $a=\hat{i}+\hat{j}+2\hat{k}$ is given by $\hat{a}=\frac{a}{|a|}=\frac{1}{\sqrt{6}}(\hat{i}+\hat{j}+2\hat{k})$
$=\frac{1}{\sqrt{6}}\hat{i}+\frac{1}{\sqrt{6}}\hat{j}+\frac{2}{\sqrt{6}}\hat{k}$