A) \[5a\]
B) \[6a\]
C) \[10a\]
D) \[9a\]
E) \[7a\]
Correct Answer: A
Solution :
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 \[R=\overrightarrow{OR}+\overrightarrow{OS}+\overrightarrow{OT}\] \[=a\left( \frac{\hat{i}+\hat{k}}{\sqrt{2}} \right)+2a\left( \frac{\hat{k}+\hat{i}}{\sqrt{2}} \right)+3a\left( \frac{\hat{i}+\hat{j}}{\sqrt{2}} \right)\] \[=\frac{5a}{\sqrt{2}}\hat{i}+\frac{4a}{\sqrt{2}}\hat{j}+\frac{3a}{\sqrt{2}}\hat{k}\] \[\therefore \] \[|R|=\sqrt{\frac{25{{a}^{2}}}{2}+\frac{16{{a}^{2}}}{2}+\frac{9{{a}^{2}}}{2}}=5a\]You need to login to perform this action.
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