Q68 / 180 · COMEDK 2020 · Mathematics
A vector perpendicular to the plane containing the points $A(1,-1,2), B(2,0,-1), C(0,2,1)$ is
- Correct answerA$8 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
- B$4 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$
- C$\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$
- D$3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
Solution
We have,
$A(1,-1,2), B(2,0,-1)$ and $C(0,2,1)$ $\therefore \quad \mathbf{A B}=\{2-1) \mathbf{i}+(0-(-1)) \mathbf{j}+(-1-2) k$ $=\mathbf{i}+\mathbf{j}-3 \mathbf{k}$ and $A C=(0-1) \mathbf{i}+(2-(-1)) \mathbf{j}+(1-2) k$ $=-\mathbf{i}+3 \mathbf{j}-\mathrm{k}$
Now, required vector $=\mathbf{A B} \times \mathbf{A C}$
$$
\begin{aligned}
&=\left|\begin{array}{ccc}
\mathrm{i} & \mathbf{j} & \mathbf{k} \\
1 & 1 & -3 \\
-1 & 3 & -1
\end{array}\right| \\
&=8 \mathrm{i}+4 \mathrm{j}+4 \mathrm{k}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.