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3D Geometry question

2025 · 22 Jan · Shift 2 · Q27
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  5. /2025 · 22 Jan · Shift 2 · Q27

3D Geometry question

2025 · 22 Jan · Shift 2 · Q27

JEE MainMathematics3D GeometryMCQ+4 / −1
Let a line pass through two distinct points P(−2,−1,3)P(-2,-1,3)P(−2,−1,3) and QQQ, and be parallel to the vector 3i^+2j^+2k^3 \hat{i}+2 \hat{j}+2 \hat{k}3i^+2j^​+2k^. If the distance of the point Q from the point R(1,3,3)\mathrm{R}(1,3,3)R(1,3,3) is 5 , then the square of the area of △PQR\triangle P Q R△PQR is equal to :
  1. A
    148
  2. B
    144
  3. C
    136
  4. D
    140
View written solutionFree

Correct answer: C

  1. Direction of the line

The line passes through P(−2,−1,3)P(-2,-1,3)P(−2,−1,3) and is parallel to the vector d⃗=3i^+2j^+2k^=(3,2,2).\vec d=3\hat i+2\hat j+2\hat k=(3,2,2).d=3i^+2j^​+2k^=(3,2,2).

So any point QQQ on this line can be written as Q=P+td⃗=(−2,−1,3)+t(3,2,2).Q=P+t\vec d=(-2,-1,3)+t(3,2,2).Q=P+td=(−2,−1,3)+t(3,2,2). Hence, Q=(−2+3t,−1+2t,3+2t).Q=(-2+3t,-1+2t,3+2t).Q=(−2+3t,−1+2t,3+2t).

  1. Use the condition QR=5QR=5QR=5

Given R(1,3,3)R(1,3,3)R(1,3,3), we compute RQ→=Q−R=(−2+3t−1,−1+2t−3,3+2t−3)=(3t−3,2t−4,2t).\overrightarrow{RQ}=Q-R=(-2+3t-1,-1+2t-3,3+2t-3)=(3t-3,2t-4,2t).RQ​=Q−R=(−2+3t−1,−1+2t−3,3+2t−3)=(3t−3,2t−4,2t).

Since QR=5QR=5QR=5, ∣(3t−3,2t−4,2t)∣=5.|(3t-3,2t-4,2t)|=5.∣(3t−3,2t−4,2t)∣=5. So, (3t−3)2+(2t−4)2+(2t)2=25.(3t-3)^2+(2t-4)^2+(2t)^2=25.(3t−3)2+(2t−4)2+(2t)2=25.

Expand: 9(t−1)2+4(t−2)2+4t2=259(t-1)^2+4(t-2)^2+4t^2=259(t−1)2+4(t−2)2+4t2=25 9(t2−2t+1)+4(t2−4t+4)+4t2=259(t^2-2t+1)+4(t^2-4t+4)+4t^2=259(t2−2t+1)+4(t2−4t+4)+4t2=25 9t2−18t+9+4t2−16t+16+4t2=259t^2-18t+9+4t^2-16t+16+4t^2=259t2−18t+9+4t2−16t+16+4t2=25 17t2−34t+25=2517t^2-34t+25=2517t2−34t+25=25 17t2−34t=017t^2-34t=017t2−34t=0 17t(t−2)=0.17t(t-2)=0.17t(t−2)=0.

Thus, t=0ort=2.t=0 \quad \text{or} \quad t=2.t=0ort=2.

  • t=0t=0t=0 gives Q=PQ=PQ=P, but PPP and QQQ are distinct, so reject it.
  • Hence t=2t=2t=2.

Therefore, Q=(−2+6,−1+4,3+4)=(4,3,7).Q=(-2+6,-1+4,3+4)=(4,3,7).Q=(−2+6,−1+4,3+4)=(4,3,7).

  1. Find the area of △PQR\triangle PQR△PQR

Take vectors from PPP: PQ→=Q−P=(4−(−2),3−(−1),7−3)=(6,4,4),\overrightarrow{PQ}=Q-P=(4-(-2),3-(-1),7-3)=(6,4,4),PQ​=Q−P=(4−(−2),3−(−1),7−3)=(6,4,4), PR→=R−P=(1−(−2),3−(−1),3−3)=(3,4,0).\overrightarrow{PR}=R-P=(1-(-2),3-(-1),3-3)=(3,4,0).PR=R−P=(1−(−2),3−(−1),3−3)=(3,4,0).

Area of triangle: Area=12∣PQ→×PR→∣.\text{Area}=\frac12\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|.Area=21​​PQ​×PR​.

Now compute the cross product:

\begin{vmatrix} \hat i & \hat j & \hat k\\ 6 & 4 & 4\\ 3 & 4 & 0 \end{vmatrix}.$$ $$=\hat i(4\cdot 0-4\cdot 4)-\hat j(6\cdot 0-4\cdot 3)+\hat k(6\cdot 4-4\cdot 3)$$ $$=(-16,12,12).$$ So, $$\left|\overrightarrow{PQ}\times\overrightarrow{PR}\right|^2=(-16)^2+12^2+12^2=256+144+144=544.$$ Hence, $$\text{Area}^2=\frac14\cdot 544=136.$$ 4. **Match with options** Thus the square of the area of $\triangle PQR$ is $$\boxed{136}.$$ So the correct option is **C**.
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