Vector Algebra Question 182
Question: Let $ \vec{a}=\hat{i}+\hat{j};\hat{b}=2\hat{i}-\hat{k} $ . Then vector $ \vec{r} $ satisfying the equations $ \vec{r}\times \vec{a}=\vec{b}\times \vec{a} $ and $ \vec{r}\times \vec{b}=\vec{a}\times \vec{b} $ is
Options:
A) $ \hat{i}-\hat{j}+\hat{k} $
B) $ 3\hat{i}-\hat{j}+\hat{k} $
C) $ 3\hat{i}+\hat{j}-\hat{k} $
D) $ \hat{i}-\hat{j}-\hat{k} $
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Answer:
Correct Answer: C
Solution:
- [c] $ \vec{r}\times \vec{a}=\vec{b}\times \vec{a} $ or $ (\vec{r}-\vec{b})\times \vec{a}=0 $ $ \vec{r}\times \vec{b}=\vec{a}\times \vec{b} $ or $ (\vec{r}-\vec{b})\times \vec{b}=0 $ If $ \vec{r}=x\hat{i}+y\hat{j}+z\hat{k} $ , then $ \begin{vmatrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ x-2 & y & z+1 \\ 1 & 1 & 0 \\ \end{vmatrix} =0 $ and $ \begin{vmatrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ x-1 & y-1 & z \\ 2 & 0 & -1 \\ \end{vmatrix} =0 $
$ \Rightarrow z+1=0,x-y=2 $ And $ y-1=0,x-1+2z=0 $
$ \Rightarrow x=3,y=1,z=-1 $
BETA
