Question

The vector $ \vec{c} $ is perpendicular to the vectors $ \vec{a}=(2,-3,1),\vec{b}=(1,-2,3) $ and satisfies the condition $ \vec{c}.(\hat{i}+2\hat{j}-7\hat{k})=10. $ Then, $ \vec{c} $ is equal to

• A. $ 7\hat{i}+5\hat{j}+\hat{k} $
• B. $ -7\hat{i}-5\hat{j}-\hat{k} $
• C. $ \hat{i}+\hat{j}-\hat{k} $
• D. $ \hat{i}+\hat{j}+\hat{k} $

Answer 1

Answer: (A)

Since, $ \vec{c} $ is perpendicular to the vectors $ \vec{a}=(2,-3,1) $ and $ \vec{b}=(1,-2,3), $ therefore $ \vec{c} $ is parallel to $ \vec{a}\times \vec{b}. $
$ \therefore $ $ \vec{c}=\lambda (\vec{a}\times \vec{b}) $
$ \Rightarrow $ $ \begin{vmatrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 2 & -3 & 1 \\ 1 & -2 & 3 \\ \end{vmatrix} $
$ \Rightarrow $ $ \vec{c}=\lambda (-7\hat{i}-5\hat{j}-\hat{k}) $
Also, it is given that
$ \vec{c}.(\hat{i}+2\hat{j}-7\hat{k})=10 $
$ \Rightarrow $ $ \lambda (-7\hat{i}-5\hat{j}-\hat{k}).(\hat{i}+2\hat{j}-7\hat{k})=10 $
$ \Rightarrow $ $ \lambda (-7-10+7)=10 $
$ \Rightarrow $ $ -10\lambda =10 $
$ \Rightarrow $ $ \lambda =-1 $
Hence, $ \vec{c}=(-1)(-7\hat{i}-5\hat{j}-\hat{k}) $
$ =(7\hat{i}+5\hat{j}+\hat{k}) $