A) \[\frac{1}{3}(\hat{i}-2\hat{j}+\hat{k})\]
B) \[\frac{1}{3}(-\,\hat{i}+2\hat{j}+5\hat{k})\]
C) \[\frac{1}{3}(\hat{i}+2\hat{j}-5\hat{k})\]
D) \[\frac{1}{3}(-\,\hat{i}+2\hat{j}-5\hat{k})\]
Correct Answer: B
Solution :
| \[\vec{a}\,\,\times \,\,\vec{b}=\vec{a}\,\,\times \,\,(\vec{a}\,\,\times \,\,\vec{c})\]\[=(\vec{a}.\,\,\vec{c})\,\,\vec{a}-(\vec{a}.\,\,\vec{a})\,\,\vec{c}=2\,\,\vec{a}-3\,\,\vec{c}\] |
| But \[\vec{a}\,\,\times \,\,\vec{b}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 1 & 1 \\ 1 & -\,2 & 1 \\ \end{matrix} \right|=3\hat{i}\,\,-\,\,3\hat{k}\] Hence \[3\,\vec{c}=2\,\vec{a}-(3\,\hat{i}-3\,\hat{k})\] |
| \[=(2\,\hat{i}+2\hat{j}+2\,\hat{k})-(3\,\hat{i}-3\,\hat{k})\] |
| \[=-\hat{i}+2\hat{j}+5\hat{k}\] \[\Rightarrow \overrightarrow{c}=\frac{1}{3}(-\,\hat{i}++2\hat{j}+5\hat{k})\] |
You need to login to perform this action.
You will be redirected in
3 sec
