The unit vector perpendicular to the plane determined by P(1,-1,2), Q(2,0,-1) and R(0,2,1) is ⋅s.

Question

The unit vector perpendicular to the plane determined by P(1,-1,2), Q(2,0,-1) and R(0,2,1) is ⋅s. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

A unit vector perpendicular to the plane determined by

P, Q, R

= \pm \frac{( PQ ) \times ( PR )}{∣ PQ \times PR ∣}

∴

Unit vector

= \pm \frac{( PQ ) \times ( PR )}{∣ PQ \times PR ∣}

where,

PQ = \hat{i} + \hat{j} - 3 \hat{k}

and

PR = \hat{i} - 3 \hat{j} - \hat{k}

∴ PQ \times PR = ∣ ∣ \hat{i} 1 - 1 \hat{j} 13 \hat{k} - 3 - 1 ∣ ∣

= \hat{i} (- 1 + 9) - \hat{j} (- 1 - 3) + \hat{k} (3 + 1)

= 8 i + 4 j + 4 \hat{k}

\Rightarrow ∣ PQ \times PR ∣ = 4 4 + 1 + 1 = 46 = \pm \frac{4 ( 2 i ^ + j ^ + k ^ )}{4 6}

= \pm \frac{( 2 i ^ + j ^ + k ^ )}{6}

Q. The unit vector perpendicular to the plane determined by P(1,−1,2),Q(2,0,−1) and R(0,2,1) is ⋯.