Question

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

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

Answer 1

The correct option is B

$8i+4j+4k$

Explanation for the correct option:

Step 1: Find the Cross product of $A$ and $B$.

Given: $A(1,-1,2)$ and $B(2,0,-1)$

So, $\begin{array}{rcl}A\times B& =& \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 2\\ 2& 0& -1\end{array}\right|\\ & =& \left(\left(-1\right)·\left(-1\right)+2·0\right)\hat{i}-\left(1·\left(-1\right)-2·2\right)\hat{j}+\left(1·0-\left(-1\right)·2\right)\hat{k}\\ & =& \hat{i}+5\hat{j}+2\hat{k}\end{array}$

Step 2: Find the Cross product of $B$ and $C$.

Given: $B(2,0,-1)$ and $C(0,2,1)$

So, $\begin{array}{rcl}B\times C& =& \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 0& -1\\ 0& 2& 1\end{array}\right|\\ & =& \left(0·1-2·\left(-1\right)\right)\hat{i}-\left(2·1-\left(-1\right)·0\right)\hat{j}+\left(2·2-0·0\right)\hat{k}\\ & =& 2\hat{i}-2\hat{j}+4\hat{k}\end{array}$

Step 3: Find the Cross product of $A$ and $C$.

Given: $A(1,-1,2)$ and $C(0,2,1)$

So, $\begin{array}{rcl}C\times A& =& \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 2& 1\\ 1& -1& 2\end{array}\right|\\ & =& \left(2·2-1·\left(-1\right)\right)\hat{i}-\left(0·2-1·1\right)\hat{j}+\left(0·\left(-1\right)-1·2\right)\hat{k}\\ & =& 5\hat{i}+\hat{j}-2\hat{k}\end{array}$

Step 4: Find the dot product of the three vectors:

Use the cross products of the three vectors taken two at a time to find the dot product.

$\begin{array}{rcl}A·B·C& =& (\hat{i}+5\hat{j}+2\hat{k})+(2\hat{i}-2\hat{j}+4\hat{k})+(5\hat{i}+\hat{j}-2\hat{k})\\ & =& 8\hat{i}+4\hat{j}+4\hat{k}\end{array}$ $\left[\because A·B·C=\left(A\times B\right)+\left(B\times C\right)+\left(C\times A\right)\right]$

Hence, option (B) is correct.