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← JEE Advanced 2024 · Paper 2
MathematicsMultiple correct+4 / -2Question 6 of 51

JEE Advanced 2024 · Paper 2, Mathematics question 6

A straight line drawn from the point P(1,3,2)P(1,3,2), parallel to the line x−21=y−42=z−61\dfrac{x-2}{1} = \dfrac{y-4}{2} = \dfrac{z-6}{1}, intersects the plane L1:x−y+3z=6L_1 : x - y + 3z = 6 at the point QQ. Another straight line which passes through QQ and is perpendicular to the plane L1L_1 intersects the plane L2:2x−y+z=−4L_2 : 2x - y + z = -4 at the point RR. Then which of the following statements is (are) TRUE?

Choose one or more options
Answer and worked solution

Official answer: A, C

Q. The line through PP is (1+t, 3+2t, 2+t)(1+t,\,3+2t,\,2+t). Substituting into x−y+3z=6x-y+3z=6:
(1+t)−(3+2t)+3(2+t)=4+2t=6  ⟹  t=1,(1+t)-(3+2t)+3(2+t)=4+2t=6\implies t=1,
so Q=(2,5,3)Q=(2,5,3) and PQ=∣t∣1+4+1=6PQ=|t|\sqrt{1+4+1}=\sqrt6.

R. The normal line from QQ is (2+s, 5−s, 3+3s)(2+s,\,5-s,\,3+3s). Substituting into 2x−y+z=−42x-y+z=-4:
2(2+s)−(5−s)+(3+3s)=2+6s=−4  ⟹  s=−1,2(2+s)-(5-s)+(3+3s)=2+6s=-4\implies s=-1,
so R=(1,6,0)R=(1,6,0) and QR=11QR=\sqrt{11}.

  • (A) PQ=6PQ=\sqrt6. ✓
  • (B) R=(1,6,0)R=(1,6,0), not (1,6,3)(1,6,3). ✗
  • (C) Centroid =(1+2+13,3+5+63,2+3+03)=(43,143,53)=\left(\frac{1+2+1}{3},\frac{3+5+6}{3},\frac{2+3+0}{3}\right)=\left(\frac43,\frac{14}3,\frac53\right). ✓
  • (D) PR=∣(0,3,−2)∣=13PR=|(0,3,-2)|=\sqrt{13}, so the perimeter is 6+11+13\sqrt6+\sqrt{11}+\sqrt{13}. ✗
Topics:Equation of a line in spaceEquation of a planeAngles and distances
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