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

A unit vector perpendicular to the plane determined by the points (1, – 1, 2), (2, 0, – 1) and (0, 2, 1) is

A

±16(2i+j+k)\pm \frac { 1 } { \sqrt { 6 } } ( 2 \mathbf { i } + \mathbf { j } + \mathbf { k } )

B

16(i+2j+k)\frac { 1 } { \sqrt { 6 } } ( \mathbf { i } + 2 \mathbf { j } + \mathbf { k } )

C

16(i+j+k)\frac { 1 } { \sqrt { 6 } } ( \mathbf { i } + \mathbf { j } + \mathbf { k } )

D

16(2i−j−k)\frac { 1 } { \sqrt { 6 } } ( 2 \mathbf { i } - \mathbf { j } - \mathbf { k } )

Answer

±16(2i+j+k)\pm \frac { 1 } { \sqrt { 6 } } ( 2 \mathbf { i } + \mathbf { j } + \mathbf { k } )

Explanation

Solution

a=i+j−3k\mathbf { a } = \mathbf { i } + \mathbf { j } - 3 \mathbf { k } b=−2i+2j+2k\mathbf { b } = - 2 \mathbf { i } + 2 \mathbf { j } + 2 \mathbf { k }

a×b=∣ijk11−3−222∣=8i+4j+4k\mathbf { a } \times \mathbf { b } = \left| \begin{array} { c c c } \mathbf { i } & \mathbf { j } & \mathbf { k } \\ 1 & 1 & - 3 \\ - 2 & 2 & 2 \end{array} \right| = 8 \mathbf { i } + 4 \mathbf { j } + 4 \mathbf { k }

Hence unit vector =±2i+j+k6= \pm \frac { 2 \mathbf { i } + \mathbf { j } + \mathbf { k } } { \sqrt { 6 } }