JEE Advanced 2025 Paper 1 Q05 Mathematics Vectors & 3D Geometry 3D Lines Hard

JEE Advanced 2025 Paper 1 · Q05 · 3D Lines

Let $L_1$ be the line of intersection of the planes $2x + 3y + z = 4$ and $x + 2y + z = 5$. Let $L_2$ be the line through $P(2, -1, 3)$ parallel to $L_1$. Let $M$ be the plane $2x + y - 2z = 6$. Let $Q$ be the point where $L_2$ meets $M$, and $R$ be the foot of perpendicular from $P$ to $M$. Which statements are TRUE?

  1. A. $|PQ| = 9\sqrt{3}$
  2. B. $|QR| = 15$
  3. C. Area of triangle $PQR = \frac{3}{2}\sqrt{234}$
  4. D. Acute angle between $PQ$ and $PR$ is $\cos^{-1}\left(\frac{1}{2\sqrt{3}}\right)$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, C

Solution

Direction of $L_1$: $\vec{n_1} \times \vec{n_2}$ where $\vec{n_1}=(2,3,1)$, $\vec{n_2}=(1,2,1)$. Cross product $= (1, -1, 1)$. $L_2$: $(2+t, -1-t, 3+t)$. Plug into $M$: $2(2+t)+(-1-t)-2(3+t) = -3 - t = 6$ $\Rightarrow$ $t = -9$. $Q = (-7, 8, -6)$. $\vec{PQ} = (-9, 9, -9)$, $|PQ| = 9\sqrt{3}$. (A) TRUE. Foot of perpendicular $R$ from $P$ along $(2,1,-2)$: $(2+2s, -1+s, 3-2s)$ on plane gives $-3+9s=6$ $\Rightarrow$ $s=1$, so $R = (4, 0, 1)$ and $\vec{PR} = (2,1,-2)$, $|PR| = 3$. $\vec{QR} = R - Q = (11, -8, 7)$, $|QR| = \sqrt{121+64+49} = \sqrt{234} \approx 15.3$, NOT $15$. (B) FALSE. Area $= \frac{1}{2}|\vec{PQ} \times \vec{PR}|$. $\vec{PQ} \times \vec{PR} = -9 \cdot ((1,-1,1)\times(2,1,-2)) = -9 \cdot (1,4,3)$; magnitude $9\sqrt{26}$. Area $= \frac{9}{2} \sqrt{26} = \frac{3}{2} \cdot 3\sqrt{26} = \frac{3}{2}\sqrt{234}$. (C) TRUE. $\vec{PQ} \cdot \vec{PR} = -18+9+18 = 9$; $\cos\theta = 9 / (9\sqrt{3} \cdot 3) = \frac{1}{3\sqrt{3}} \ne \frac{1}{2\sqrt{3}}$. (D) FALSE. Answer: A, C.

Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →