question_answer

Let ABCD be the parallelogram whose sides AB and AD are represented by the vectors \[2\hat{i}+4\hat{j}-5\hat{k}\]and\[\hat{i}+2\hat{j}+3\hat{k}\] respectively. Then, if a is a unit vector parallel to AC then a equals:

A)  \[\frac{1}{3}(3\hat{i}-6\hat{j}-2\hat{k})\]

B)  \[\frac{1}{3}(3\hat{i}+6\hat{j}+2\hat{k})\]

C)  \[\frac{1}{3}(3\hat{i}-6\hat{j}-3\hat{k})\]

D)  \[\frac{1}{7}(3\hat{i}+6\hat{j}-2\hat{k})\]

E)  \[\frac{1}{7}(3\hat{i}+5\hat{j}-3\hat{k})\]

Correct Answer: D

Solution :

Let \[{{\overrightarrow{R}}_{1}}=2\hat{i}+4\hat{j}-5\hat{k}\] and        \[{{\overrightarrow{R}}_{2}}=\hat{i}+2\hat{j}+3\hat{k}\] \[\therefore \]\[\overrightarrow{R}\](along \[\overset{\to }{\mathop{AC}}\,\]) \[={{\overrightarrow{R}}_{1}}+{{\overrightarrow{R}}_{2}}\]                 \[=3\hat{i}+6\hat{j}-2\hat{k}\] \[\therefore \] \[\overrightarrow{a}\] (unit vector along\[\overset{\to }{\mathop{AC}}\,\])\[=\frac{\overset{\to }{\mathop{R}}\,}{|\overset{\to }{\mathop{R}}\,|}\] \[=\frac{3i+6\hat{j}-2\hat{k}}{\sqrt{9+36+4}}\]                 \[=\frac{1}{7}(3\hat{i}+6\hat{j}-2\hat{k})\]