Question

The unit vectors perpendicular to the plane determined by the points $A(1,-1,2)$, $B(2,0,-1)$, $C(0,2,1)$ is

• A. $\pm(\frac{3\hat{i}+\hat{j}+\hat{k}}{\sqrt{11}})$
• B. $\pm(\frac{-\hat{i}+2\hat{j}+\hat{k}}{\sqrt{6}})$
• C. $\pm(\frac{2\hat{i}+\hat{j}+\hat{k}}{\sqrt{6}})$
• D. $\pm(\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}})$

Hint:

You can simplify the cross product vector by dividing by a common factor before normalizing.

Answer 1

The Correct Option is C

Step 1: Concept
Unit normal vector $\hat{n} = \pm \frac{\vec{AB} \times \vec{AC}}{|\vec{AB} \times \vec{AC}|}$.

Step 2: Analysis
$\vec{AB} = (1, 1, -3)$.
$\vec{AC} = (-1, 3, -1)$.

Step 3: Calculation
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -3 \\ -1 & 3 & -1 \end{vmatrix} = \hat{i}(8) - \hat{j}(-4) + \hat{k}(4) = (8, 4, 4) \sim (2, 1, 1)$.
Magnitude of $(2, 1, 1) = \sqrt{4+1+1} = \sqrt{6}$.

Step 4: Conclusion
Hence, the unit vectors are $\pm \frac{2\hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}$. Final Answer: (C)