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Question: If \(A ( - 1,2,3 ) , B ( 1,1,1 )\) and \(C ( 2 , - 1,3 )\) are points on a plane. A unit normal ve...

If A(−1,2,3),B(1,1,1)A ( - 1,2,3 ) , B ( 1,1,1 ) and C(2,−1,3)C ( 2 , - 1,3 ) are points on a plane. A unit normal vector to the plane ABC is

A

±(2i+2j+k3)\pm \left( \frac { 2 \mathbf { i } + 2 \mathbf { j } + \mathbf { k } } { 3 } \right)

B

±(2i−2j+k3)\pm \left( \frac { 2 \mathbf { i } - 2 \mathbf { j } + \mathbf { k } } { 3 } \right)

C

±(2i−2j−k3)\pm \left( \frac { 2 \mathbf { i } - 2 \mathbf { j } - \mathbf { k } } { 3 } \right)

D

−(2i+2j+k3)- \left( \frac { 2 \mathbf { i } + 2 \mathbf { j } + \mathbf { k } } { 3 } \right)

Answer

±(2i+2j+k3)\pm \left( \frac { 2 \mathbf { i } + 2 \mathbf { j } + \mathbf { k } } { 3 } \right)

Explanation

Solution

AB→=2i−j−2k\overrightarrow { A B } = 2 \mathbf { i } - \mathbf { j } - 2 \mathbf { k } AC→=3i−3j+0k\overrightarrow { A C } = 3 \mathbf { i } - 3 \mathbf { j } + 0 \mathbf { k }

AB→×AC→=∣ijk2−1−23−30∣=(−6i−6j−3k)\overrightarrow { A B } \times \overrightarrow { A C } = \left| \begin{array} { c c c } \mathbf { i } & \mathbf { j } & \mathbf { k } \\ 2 & - 1 & - 2 \\ 3 & - 3 & 0 \end{array} \right| = ( - 6 \mathbf { i } - 6 \mathbf { j } - 3 \mathbf { k } )

Hence unit vector =±(2i+2j+k3)= \pm \left( \frac { 2 \mathbf { i } + 2 \mathbf { j } + \mathbf { k } } { 3 } \right)