JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Three Dimensional Geometry

1. (A)The length of the line segment $PQ$ is $\sqrt{6}$✓ Correct
2. (B)The coordinates of $R$ are $(1,6,3)$
3. (C)The centroid of the triangle $PQR$ is $\left(\frac{4}{3},\frac{14}{3},\frac{5}{3}\right)$✓ Correct
4. (D)The perimeter of the triangle $PQR$ is $\sqrt{2}+\sqrt{6}+\sqrt{11}$

Answer: (A), (C)

Solution

The line through $P(1,3,2)$ parallel to the given line has direction ratios $(1,2,1)$:

$$\frac{x-1}{1}=\frac{y-3}{2}=\frac{z-2}{1}=\lambda$$

Substituting into $L_1:x-y+3z=6$:

$$(\lambda +1)-(2\lambda +3)+3(\lambda +2)=6\Rightarrow 2\lambda =2\Rightarrow \lambda =1$$

So $Q=(2,5,3)$ and $PQ=\sqrt{1+4+1}=\sqrt{6}$. ✓(A)

The line through $Q$ perpendicular to $L_1$ has direction ratios $(1,-1,3)$:

$$\frac{x-2}{1}=\frac{y-5}{-1}=\frac{z-3}{3}=t$$

Substituting into $L_2:2x-y+z=-4$:

$$2(t+2)-(5-t)+(3t+3)=-4\Rightarrow 6t=-6\Rightarrow t=-1$$

So $R=(1,6,0)$. ✗(B) states $(1,6,3)$

Centroid: $\left(\frac{1+2+1}{3},\frac{3+5+6}{3},\frac{2+3+0}{3}\right)=\left(\frac{4}{3},\frac{14}{3},\frac{5}{3}\right)$. ✓(C)

Perimeter: $PQ=\sqrt{6}$, $QR=\sqrt{1+1+9}=\sqrt{11}$, $PR=\sqrt{0+9+4}=\sqrt{13}$.
Perimeter = $\sqrt{6}+\sqrt{11}+\sqrt{13}$. ✗(D)

Correct answers: (A), (C)

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