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3D Geometry question

2024 · Shift 2 · Q23
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3D Geometry question

2024 · Shift 2 · Q23

JEE AdvancedMathematics3D GeometryMultiple correct+4 / −2
A straight line drawn from the point P(1,3,2)P(1,3,2)P(1,3,2), parallel to the line x−21=y−42=z−61\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}1x−2​=2y−4​=1z−6​, intersects the plane L1:x−y+3z=6L_1: x-y+3 z=6L1​:x−y+3z=6 at the point QQQ. Another straight line which passes through QQQ and is perpendicular to the plane L1L_1L1​ intersects the plane L2:2x−y+z=−4L_2: 2 x-y+z=-4L2​:2x−y+z=−4 at the point RRR. Then which of the following statements is (are) TRUE?
  1. A
    The length of the line segment PQP QPQ is 6\sqrt{6}6​
  2. B
    The coordinates of RRR are (1,6,3)(1,6,3)(1,6,3)
  3. C
    The centroid of the triangle PQRP Q RPQR is (43,143,53)\left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right)(34​,314​,35​)
  4. D
    The perimeter of the triangle PQRP Q RPQR is 2+6+11\sqrt{2}+\sqrt{6}+\sqrt{11}2​+6​+11​
View written solutionFree

Correct answer: A, C

  1. Find the line through P(1,3,2)P(1,3,2)P(1,3,2) parallel to x−21=y−42=z−61\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}1x−2​=2y−4​=1z−6​

    The given line has direction ratios d⃗=(1,2,1).\vec d=(1,2,1).d=(1,2,1).

    So the required line through P(1,3,2)P(1,3,2)P(1,3,2) is x=1+t,y=3+2t,z=2+t.x=1+t,\quad y=3+2t,\quad z=2+t.x=1+t,y=3+2t,z=2+t.

  2. Find point QQQ where this line meets plane L1:x−y+3z=6L_1:x-y+3z=6L1​:x−y+3z=6

    Substitute the parametric coordinates into the plane equation: (1+t)−(3+2t)+3(2+t)=6.(1+t)-(3+2t)+3(2+t)=6.(1+t)−(3+2t)+3(2+t)=6.

    Simplify: 1+t−3−2t+6+3t=61+t-3-2t+6+3t=61+t−3−2t+6+3t=6 4+2t=64+2t=64+2t=6 2t=2  ⟹  t=1.2t=2\implies t=1.2t=2⟹t=1.

    Therefore, Q=(1+1, 3+2, 2+1)=(2,5,3).Q=(1+1,\,3+2,\,2+1)=(2,5,3).Q=(1+1,3+2,2+1)=(2,5,3).

  3. Check option A: length PQPQPQ

    PQ→=Q−P=(2−1,5−3,3−2)=(1,2,1).\overrightarrow{PQ}=Q-P=(2-1,5-3,3-2)=(1,2,1).PQ​=Q−P=(2−1,5−3,3−2)=(1,2,1).

    Hence, PQ=12+22+12=6.PQ=\sqrt{1^2+2^2+1^2}=\sqrt{6}.PQ=12+22+12​=6​.

    So A is true.

  4. Find the line through QQQ perpendicular to plane L1L_1L1​

    Plane L1:x−y+3z=6L_1:x-y+3z=6L1​:x−y+3z=6 has normal vector n⃗1=(1,−1,3).\vec n_1=(1,-1,3).n1​=(1,−1,3).

    A line perpendicular to L1L_1L1​ through Q(2,5,3)Q(2,5,3)Q(2,5,3) therefore has parametric form x=2+s,y=5−s,z=3+3s.x=2+s,\quad y=5-s,\quad z=3+3s.x=2+s,y=5−s,z=3+3s.

  5. Find point RRR where this line meets plane L2:2x−y+z=−4L_2:2x-y+z=-4L2​:2x−y+z=−4

    Substitute into L2L_2L2​: 2(2+s)−(5−s)+(3+3s)=−4.2(2+s)-(5-s)+(3+3s)=-4.2(2+s)−(5−s)+(3+3s)=−4.

    Simplify: 4+2s−5+s+3+3s=−44+2s-5+s+3+3s=-44+2s−5+s+3+3s=−4 2+6s=−42+6s=-42+6s=−4 6s=−6  ⟹  s=−1.6s=-6\implies s=-1.6s=−6⟹s=−1.

    So, R=(2−1, 5−(−1), 3+3(−1))=(1,6,0).R=(2-1,\,5-(-1),\,3+3(-1))=(1,6,0).R=(2−1,5−(−1),3+3(−1))=(1,6,0).

  6. Check option B

    Option B says R=(1,6,3)R=(1,6,3)R=(1,6,3), but we found R=(1,6,0).R=(1,6,0).R=(1,6,0).

    So B is false.

  7. Check option C: centroid of triangle PQRPQRPQR

    P=(1,3,2),Q=(2,5,3),R=(1,6,0).P=(1,3,2),\quad Q=(2,5,3),\quad R=(1,6,0).P=(1,3,2),Q=(2,5,3),R=(1,6,0).

    Centroid is

    =\left(\frac{4}{3},\frac{14}{3},\frac{5}{3}\right).$$ This matches option C. So **C is true**.

  8. Check option D: perimeter of triangle PQRPQRPQR

    We already have PQ=6.PQ=\sqrt{6}.PQ=6​.

    Next,

    =\sqrt{1+1+9}=\sqrt{11}.$$ Also, $$PR=\sqrt{(1-1)^2+(6-3)^2+(0-2)^2} =\sqrt{0+9+4}=\sqrt{13}.$$ Therefore perimeter is $$PQ+QR+PR=\sqrt6+\sqrt{11}+\sqrt{13}.$$ Option D gives $\sqrt2+\sqrt6+\sqrt{11}$, which is incorrect. So **D is false**.

  9. Final conclusion

    The true statements are: A, C\boxed{A,\ C}A, C​

  10. Comparison with stored correct answer

Stored correct answer: A, C

Our derived answer matches the stored answer.

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