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Question: A unit vector perpendicular to the plane determined by the points \(P(1, - 1,2),Q(2,0, - 1)\) and \(...

A unit vector perpendicular to the plane determined by the points P(1,−1,2),Q(2,0,−1)P(1, - 1,2),Q(2,0, - 1) and R(0,2,1)R(0,2,1) is

A

2i−j+k6\frac{2\mathbf{i} - \mathbf{j} + \mathbf{k}}{\sqrt{6}}

B

2i+j+k6\frac{2\mathbf{i} + \mathbf{j} + \mathbf{k}}{\sqrt{6}}

C

−2i+j+k6\frac{- 2\mathbf{i} + \mathbf{j} + \mathbf{k}}{\sqrt{6}}

D

2i+j−k6\frac{2\mathbf{i} + \mathbf{j} - \mathbf{k}}{\sqrt{6}}

Answer

2i+j+k6\frac{2\mathbf{i} + \mathbf{j} + \mathbf{k}}{\sqrt{6}}

Explanation

Solution

A vector perpendicular to the plane determined by the points P(1,−1,2);P(1, - 1,2); Q(2,0,−1)Q(2,0, - 1) and R(0,2,1)R(0,2,1) is given by

QR→×PR→⇒(−2i+2j+2k)×(−i+3j−k)\overset{\rightarrow}{QR} \times \overset{\rightarrow}{PR} \Rightarrow ( - 2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) \times ( - \mathbf{i} + 3\mathbf{j} - \mathbf{k})

Therefore, unit vector =2i+j+k4+1+1=2i+j+k6.= \frac{2\mathbf{i} + \mathbf{j} + \mathbf{k}}{\sqrt{4 + 1 + 1}} = \frac{2\mathbf{i} + \mathbf{j} + \mathbf{k}}{\sqrt{6}}.