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 $\vec{a}\times \vec{b}.$
$\therefore$ $\vec{c}=\lambda (\vec{a}\times \vec{b})$
$\Rightarrow$ $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -3& 1\\ 1& -2& 3\end{array}\right|$
$\Rightarrow$ $\vec{c}=\lambda (-7\hat{i}-5\hat{j}-\hat{k})$
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, $\vec{c}=(-1)(-7\hat{i}-5\hat{j}-\hat{k})$
$=(7\hat{i}+5\hat{j}+\hat{k})$