Question

A vector perpendicular to the plane containing the points $A(1,-1,2),B(2,0,-1),C(0,2,1)$ is

• A. $8\hat{i}+4\hat{j}+4\hat{k}$
• B. $4\hat{i}+8\hat{j}-4\hat{k}$
• C. $\hat{i}+\hat{j}-\hat{k}$
• D. $3\hat{i}+\hat{j}+2\hat{k}$

Answer 1

Answer: (A)

We know that a vector perpendicular to the plane containing the points
$A,B,C$ is given by $A\times B+B\times C+C\times A$
We have, $A=\hat{i}-\hat{j}+2\hat{k},\vec{B}=2\hat{i}+0\hat{j}-\hat{k}$
and $C=0\hat{i}+2\hat{j}+\hat{k}$
Now, $A\times B=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 2\\ 2& 0& -1\end{array}\right|=\hat{i}+5\hat{j}+2\hat{k}$
$B\times C=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 0& -1\\ 0& 2& 1\end{array}\right|=2\hat{i}-2\hat{j}+4\hat{k}$
$C\times \vec{A}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 2& 1\\ 1& -1& 2\end{array}\right|=5\hat{i}+\hat{j}-2\hat{k}$
Thus, $A\times B+B\times C+C\times B=(\hat{i}+5\hat{j}+2\hat{k})$
$+(2\hat{i}-2\hat{j}+4\hat{k})+(5\hat{i}+\hat{j}-2\hat{k})$
$=8\hat{i}+4\hat{j}+4\hat{k}$