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MathematicsSingle correct+3 / -1Question 49 of 60

IIT-JEE 2012 · Paper 1, Mathematics question 49

The point PP is the intersection of the straight line joining the points Q(2,3,5)Q(2,3,5) and R(1,−1,4)R(1,-1,4) with the plane 5x−4y−z=15x-4y-z=1. If SS is the foot of the perpendicular drawn from the point T(2,1,4)T(2,1,4) to QRQR, then the length of the line segment PSPS is

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Answer and worked solution

Official answer: A

Line QR. (2,3,5)+s(−1,−4,−1)(2,3,5)+s(-1,-4,-1).

P (meets the plane). 5(2−s)−4(3−4s)−(5−s)=1  ⟹  12s−7=1  ⟹  s=235(2-s)-4(3-4s)-(5-s)=1\implies12s-7=1\implies s=\frac23:
P=(43,13,133).P=\left(\tfrac43,\tfrac13,\tfrac{13}3\right).

S (foot from T). QT→=(0,−2,−1)\overrightarrow{QT}=(0,-2,-1), and QT→⋅(−1,−4,−1)=9\overrightarrow{QT}\cdot(-1,-4,-1)=9 with ∣(−1,−4,−1)∣2=18|(-1,-4,-1)|^2=18, so s=12s=\frac12:
S=(32,1,92).S=\left(\tfrac32,1,\tfrac92\right).

Length. PS→=(16,23,16)\overrightarrow{PS}=\left(\frac16,\frac23,\frac16\right), ∣PS∣=1+16+136=12|PS|=\sqrt{\frac{1+16+1}{36}}=\frac1{\sqrt2}.

(A)

Topics:Equation of a line in spaceAngles and distancesEquation of a plane
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