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3D Geometry question

2025 · 7 Apr · Shift 2 · Q39
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3D Geometry question

2025 · 7 Apr · Shift 2 · Q39

JEE MainMathematics3D GeometryMCQ+4 / −1
Consider the lines L1: x - 1 = y - 2 = z and L2: x - 2 = y = z - 1. Let the feet of the perpendiculars from the point P(5, 1, -3) on the lines L1 and L2 be Q and R respectively. If the area of the triangle PQR is A, then 4A2 is equal to :
  1. A
    151
  2. B
    147
  3. C
    139
  4. D
    143
View written solutionFree

Correct answer: B

  1. Write the lines in parametric form

For L1:x−1=y−2=zL_1: x-1=y-2=zL1​:x−1=y−2=z let the common value be ttt. Then x=1+t,y=2+t,z=tx=1+t,\quad y=2+t,\quad z=tx=1+t,y=2+t,z=t So a point on L1L_1L1​ is Q=(1+t,2+t,t)Q=(1+t,2+t,t)Q=(1+t,2+t,t) and its direction vector is d⃗1=(1,1,1).\vec d_1=(1,1,1).d1​=(1,1,1).

For L2:x−2=y=z−1L_2: x-2=y=z-1L2​:x−2=y=z−1 let the common value be sss. Then x=2+s,y=s,z=1+sx=2+s,\quad y=s,\quad z=1+sx=2+s,y=s,z=1+s So a point on L2L_2L2​ is R=(2+s,s,1+s)R=(2+s,s,1+s)R=(2+s,s,1+s) and its direction vector is d⃗2=(1,1,1).\vec d_2=(1,1,1).d2​=(1,1,1).

Thus both lines are parallel.

  1. Find foot of perpendicular from P(5,1,−3)P(5,1,-3)P(5,1,−3) to L1L_1L1​

Since QQQ is the foot, we need PQ→⊥d⃗1.\overrightarrow{PQ}\perp \vec d_1.PQ​⊥d1​.

Now PQ→=Q−P=(1+t−5, 2+t−1, t+3)=(t−4,t+1,t+3).\overrightarrow{PQ}=Q-P=(1+t-5,\,2+t-1,\,t+3)=(t-4,t+1,t+3).PQ​=Q−P=(1+t−5,2+t−1,t+3)=(t−4,t+1,t+3).

So PQ→⋅(1,1,1)=0\overrightarrow{PQ}\cdot (1,1,1)=0PQ​⋅(1,1,1)=0 (t−4)+(t+1)+(t+3)=0(t-4)+(t+1)+(t+3)=0(t−4)+(t+1)+(t+3)=0 3t=0⇒t=0.3t=0 \Rightarrow t=0.3t=0⇒t=0.

Hence Q=(1,2,0).Q=(1,2,0).Q=(1,2,0).

  1. Find foot of perpendicular from PPP to L2L_2L2​

Similarly, PR→=R−P=(2+s−5, s−1, 1+s+3)=(s−3,s−1,s+4).\overrightarrow{PR}=R-P=(2+s-5,\,s-1,\,1+s+3)=(s-3,s-1,s+4).PR=R−P=(2+s−5,s−1,1+s+3)=(s−3,s−1,s+4).

Since RRR is the foot, PR→⋅(1,1,1)=0\overrightarrow{PR}\cdot (1,1,1)=0PR⋅(1,1,1)=0 (s−3)+(s−1)+(s+4)=0(s-3)+(s-1)+(s+4)=0(s−3)+(s−1)+(s+4)=0 3s=0⇒s=0.3s=0 \Rightarrow s=0.3s=0⇒s=0.

Hence R=(2,0,1).R=(2,0,1).R=(2,0,1).

  1. Find area of triangle PQRPQRPQR

Take vectors from PPP: PQ→=Q−P=(1−5,2−1,0+3)=(−4,1,3),\overrightarrow{PQ}=Q-P=(1-5,2-1,0+3)=(-4,1,3),PQ​=Q−P=(1−5,2−1,0+3)=(−4,1,3), PR→=R−P=(2−5,0−1,1+3)=(−3,−1,4).\overrightarrow{PR}=R-P=(2-5,0-1,1+3)=(-3,-1,4).PR=R−P=(2−5,0−1,1+3)=(−3,−1,4).

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

Compute the cross product:

=\begin{vmatrix} \hat i & \hat j & \hat k\\ -4 & 1 & 3\\ -3 & -1 & 4 \end{vmatrix}$$ $$=\hat i(1\cdot 4-3\cdot(-1)) - \hat j((-4)\cdot 4-3\cdot(-3)) + \hat k((-4)\cdot(-1)-1\cdot(-3))$$ $$=7\hat i+7\hat j+7\hat k.$$ Therefore, $$\left|\overrightarrow{PQ}\times \overrightarrow{PR}\right|=\sqrt{7^2+7^2+7^2}=7\sqrt3.$$ So $$A=\frac{7\sqrt3}{2}.$$ Then $$4A^2=4\left(\frac{7\sqrt3}{2}\right)^2=4\cdot \frac{147}{4}=147.$$ --- 5. **Match with options** $$4A^2=147$$ So the correct option is **B**. --- 6. **Comparison with stored answer** Stored correct answer: **B** Our derived answer: **B** They agree.
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