Question

The vectors of magnitude $a, 2a, 3a$ meet at a point and their directions are along the diagonals of three adjacent faces of a cube. Then, the magnitude of their resultant is

• A. $5a$
• B. $6a$
• C. $10a$
• D. $9a$

Answer 1

Answer: (A)

$5a$

Answer 2

Suppose that the sides of cube are unity and unit vector along OA, OB and OC are $\hat{i},\text{ }\hat{j},\text{ }\hat{k}$ respectively. OR, OS, OT are diagonals of cube having corresponding vector a, 2a and 3a (Magnitude) respectively.
$\therefore$ Unit vector along $OR=\frac{\hat{j}+\hat{k}}{\sqrt{2}}$
$\therefore$ Vector along $\overrightarrow{OR}=a\left( \frac{\hat{j}+\hat{k}}{\sqrt{2}} \right)$ Similarly, vector along $\overrightarrow{OS}=2a\left( \frac{\hat{k}+\hat{i}}{\sqrt{2}} \right)$ and vector along $\overrightarrow{OT}=3a\left( \frac{\hat{i}+\hat{j}}{\sqrt{2}} \right)$
$\therefore$ Resultant $R=\overrightarrow{OR}+\overrightarrow{OS}+\overrightarrow{OT}$
$=a\left( \frac{\hat{i}+\hat{k}}{\sqrt{2}} \right)+2a\left( \frac{\hat{k}+\hat{i}}{\sqrt{2}} \right)+3a\left( \frac{\hat{i}+\hat{j}}{\sqrt{2}} \right)$
$=\frac{5a}{\sqrt{2}}\hat{i}+\frac{4a}{\sqrt{2}}\hat{j}+\frac{3a}{\sqrt{2}}\hat{k}$
$\therefore$ $|R|=\sqrt{\frac{25{{a}^{2}}}{2}+\frac{16{{a}^{2}}}{2}+\frac{9{{a}^{2}}}{2}}=5a$