Question

If three vectors along coordinate axis represents the adjacent sides of a cube of length b, then the unit vector along its diagonal passing through the origin will be

• A. $\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{2}}$
• B. $\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{36}}$
• C. $\hat{i}+\hat{j}+\hat{k}$
• D. $\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$

Answer 1

Answer: (D)

$\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$

Point diagonal line$(0,0,0)$ and $(b,b,b)$

Vector along diagonal, $\overrightarrow{B}=(b-o)\hat{i}+(b-o)\hat{j}+(b-o)\hat{k}=b\hat{i}+b\hat{j}+b\hat{k}$

Unit vector along diagonal, $\hat{B}=\frac{\overrightarrow{B}}{\left|\overrightarrow{B}\right|}=\frac{b\hat{i}+b\hat{j}+b\hat{k}}{\sqrt{b^2+b^2+b^2}}=\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$

Hence, unit vector along diagonal is $\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$