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

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

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Answer
Explanation

Solution

We know that, PQ→×PR→∣PQ→×PR→∣\frac { \overrightarrow { P Q } \times \overrightarrow { P R } } { | \overrightarrow { P Q } \times \overrightarrow { P R } | }

PQ→=i+j−3k\overrightarrow { \mathrm { PQ } } = \mathrm { i } + \mathrm { j } - 3 \mathrm { k } ,

PQ→×PR→=∣ijk11−3−13−1∣=8i+4j+4k\overrightarrow { P Q } \times \overrightarrow { P R } = \left| \begin{array} { c c c } \mathbf { i } & \mathbf { j } & \mathbf { k } \\ 1 & 1 & - 3 \\ - 1 & 3 & - 1 \end{array} \right| = 8 \mathbf { i } + 4 \mathbf { j } + 4 \mathbf { k } and ∣PQ→×PR→∣=46| \overrightarrow { P Q } \times \overrightarrow { P R } | = 4 \sqrt { 6 }

Hence, the unit vector is i.e.