If position vectors of points P,Q,R with respect to origin O are overset arrow r1=3 hati-2 hatj- hatk, overset arrow r2= hati+3 hatj+4 hatk and overset arrow r3=2 hati+ hatj-2 hatk , then the distance between P and plane OQR is:-

Question

If position vectors of points P,Q,R with respect to origin O are overset arrow r1=3 hati-2 hatj- hatk, overset arrow r2= hati+3 hatj+4 hatk and overset arrow r3=2 hati+ hatj-2 hatk , then the distance between P and plane OQR is:- (A) 1 (B) 2 (C) 3 (D) None

Tardigrade Admin · Accepted Answer

Here, coordinates of P(.3,-2,-1.),Q(.1,3,4.),R(.2,1,-2.) . Equation of plane using three point form i.e. | x-0 y-0 z-0 1-0 3-0 4-0 2-0 1-0 -2-0 |=0 i.e. 2x-2y+z=0 . Now, distance of P from this plane =(|2 (. 3 .) - 2 (. - 2 .) + (. - 1 .)|/√22 + (. - 2 .)2 + 12)=3 units

Q. If position vectors of points $P, Q, R$ with respect to origin $O$ are $r_{1} \to = 3 \hat{i} - 2 \hat{j} - \hat{k}, r_{2} \to = \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $r_{3} \to = 2 \hat{i} + \hat{j} - 2 \hat{k}$ , then the distance between $P$ and plane $OQR$ is:-

$1$

$2$

$3$

None

Q. If position vectors of points P,Q,R with respect to origin O are r1​→​=3i^−2j^​−k^,r2​→​=i^+3j^​+4k^ and r3​→​=2i^+j^​−2k^ , then the distance between P and plane OQR is:-

1

2

3

None

Here, coordinates of P(3,−2,−1),Q(1,3,4),R(2,1,−2) . Equation of plane using three point form i.e. ∣∣​x−01−02−0​y−03−01−0​z−04−0−2−0​∣∣​=0 i.e. 2x−2y+z=0 . Now, distance of P from this plane =22+(−2)2+12​∣2(3)−2(−2)+(−1)∣​=3 units

Q. If position vectors of points $P, Q, R$ with respect to origin $O$ are $r_{1} \to = 3 \hat{i} - 2 \hat{j} - \hat{k}, r_{2} \to = \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $r_{3} \to = 2 \hat{i} + \hat{j} - 2 \hat{k}$ , then the distance between $P$ and plane $OQR$ is:-

$1$

$2$

$3$