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JEE Advanced 2025Paper 1
Exam
JEE Advanced
Year
2025
Paper
Paper 1
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Advanced 2025 · Paper 1 · Mathematics · Three Dimensional Geometry

Let L_1 be the line of intersection of the planes given by the...

Let L1L_1 be the line of intersection of the planes given by the equation- 2x+3y+z=4 and x+2y+z=5.2 x+3 y+z=4 \text{ and } x+2 y+z=5 . Let L2L_2 be the line passing through the point P(2,−1,3)P(2,-1,3) and parallel to L1L_1. Let MM denote the plane given by the equation- 2x+y−2z=62 x+y-2 z=6 Suppose that the line L2L_2 meets the plane MM at the point QQ. Let RR be the foot of the perpendicular drawn from PP to the plane MM. Then which of the following statements is (are) TRUE?Let L_1 be the line of intersection of the planes given by the equation- 2 x+3 y+z=4 and x+2 y+z=5 . Let L_2 be the

Options

  1. AThe length of the line segment PQP Q is 939 \sqrt{3}
  2. BThe length of the line segment QRQ R is 15
  3. CThe area of △PQR\triangle P Q R is 32234\frac{3}{2} \sqrt{234}
  4. DThe acute angle between the line segments PQandP Q andP Risis\cos ^{-1}\left(\frac{1}{2 \sqrt{3}}\right)

Correct answer

A. The length of the line segment PQP Q is 939 \sqrt{3}

Step-by-step solution

Let L_1: \vec{r}=\vec{a}+t \vec{b} \overrightarrow{\mathrm{b}}=\left|i^j^k^231121\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & 1 \\ 1 & 2 & 1\end{array}\right|=\hat{\mathrm{i}}(1)-\hat{\mathrm{j}}(1)+\hat{\mathrm{k}}(1) Dr's of L1:<1,−1,1>L_1: <1,-1,1>L1:x+11=y−0−1=z−61L2:x−21=y+1−1=z−31=λM:2x+y−2z−6=0\begin{aligned} & L_1: \frac{x+1}{1}=\frac{y-0}{-1}=\frac{z-6}{1} \\ & & L_2: \frac{x-2}{1}=\frac{y+1}{-1}=\frac{z-3}{1}=\lambda \\ & M: 2 x+y-2 z-6=0 \end{aligned}Let point on $\mathrm{L}_2(\lambda+2,-\lambda-1, \lambda+3)\begin{aligned} & 2(\lambda+2)-\lambda-1-2 \lambda-6-6=0 \\ & 2 \lambda+4-3 \lambda-13=0 \\ & \lambda=-9 \\ & \therefore Q(-7,8,-6) \end{aligned}$ <img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/J_-cnHbNEsFw31b7TfQK1lk9TvbcJFMDBjMc2Ih_VAA.original.fullsize.png"><br>Line PR: $\frac{x-2}{2}=\frac{y+1}{1}=\frac{z-3}{-2}=\mu \mathrm{R}(2 \mu+2, \mu-1,-2 \mu+3)$ Put in plane\begin{aligned} & 2(2 \mu+2)+\mu-1-2(-2 \mu+3)-6=0 \\ & 4 \mu+4+\mu-1+4 \mu-6-6=0 \\ & 9 \mu-9=0 \Rightarrow \mu=1 \\ & \mathrm{R}(4,0,1) \\ & \mathrm{P}(2,-1,3) & \mathrm{Q}(-7,8,-6) \\ & \mathrm{PQ}=\sqrt{81+81+81}=9 \sqrt{3} \\ & \mathrm{Q}(-7,8,-6) & \mathrm{R}(4,0,1) \end{aligned}\mathrm{QR}=\sqrt{121+64+49}=\sqrt{234}$ Area ( $\triangle \mathrm{PQR}$ )\begin{aligned} & =\frac{1}{2}|\overrightarrow{\mathrm{QP}} \times \overrightarrow{\mathrm{QR}}| \\ & =\frac{1}{2}\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 9 & -9 & 9 \\ 11 & -8 & 7 \end{array}\right| \\ & =\frac{3}{2} \sqrt{234} \\ & \overrightarrow{\mathrm{PQ}}=-9 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}-9 \hat{\mathrm{k}}=-9(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & \overrightarrow{\mathrm{PR}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}} \end{aligned}$$\begin{aligned} & \cos \theta=\left|\frac{\overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{PR}}}{\mathrm{PQ} \cdot \mathrm{PR}}\right| \\ & =\frac{9}{9 \sqrt{3} \times 3}=\frac{1}{3 \sqrt{3}} \\ & \theta=\cos ^{-1}\left(\frac{1}{3 \sqrt{3}}\right)\end{aligned}$
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