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3D Geometry question

2025 · 2 Apr · Shift 1 · Q27
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  5. /2025 · 2 Apr · Shift 1 · Q27

3D Geometry question

2025 · 2 Apr · Shift 1 · Q27

JEE MainMathematics3D GeometryMCQ+4 / −1
Let the vertices Q and R of the triangle PQR lie on the line x+35=y−12=z+43,QR=5\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}, \mathrm{QR}=55x+3​=2y−1​=3z+4​,QR=5 and the coordinates of the point PPP be (0,2,3)(0,2,3)(0,2,3). If the area of the triangle PQRP Q RPQR is mn\frac{m}{n}nm​ then :
  1. A
    2 m−521n=02 \mathrm{~m}-5 \sqrt{21} \mathrm{n}=02 m−521​n=0
  2. B
    m−521n=0\mathrm{m}-5 \sqrt{21} \mathrm{n}=0m−521​n=0
  3. C
    5 m−212n=05 \mathrm{~m}-21 \sqrt{2} \mathrm{n}=05 m−212​n=0
  4. D
    5 m−221n=05 \mathrm{~m}-2 \sqrt{21} \mathrm{n}=05 m−221​n=0
View written solutionFree

Correct answer: A

  1. Interpret the given line

The line is

x+35=y−12=z+43=t.\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=t.5x+3​=2y−1​=3z+4​=t.

So its parametric form is

x=−3+5t,y=1+2t,z=−4+3t.x=-3+5t,\qquad y=1+2t,\qquad z=-4+3t.x=−3+5t,y=1+2t,z=−4+3t.

A point on the line is A(−3,1,−4),A(-3,1,-4),A(−3,1,−4), and its direction vector is d⃗=(5,2,3).\vec d=(5,2,3).d=(5,2,3).

Since both QQQ and RRR lie on this line and QR=5QR=5QR=5, the base of triangle PQRPQRPQR is a segment of length 555 on this line.

  1. Use area formula with base and perpendicular distance

Area of triangle:

Δ=12×QR×distance of P from the line QR.\Delta = \frac12 \times QR \times \text{distance of }P\text{ from the line }QR.Δ=21​×QR×distance of P from the line QR.

Since QRQRQR lies on the given line, we only need the perpendicular distance from P=(0,2,3)P=(0,2,3)P=(0,2,3) to the line.

Thus,

Area=12⋅5⋅d,\text{Area} = \frac12 \cdot 5 \cdot d,Area=21​⋅5⋅d,

where ddd is the distance from PPP to the line.

  1. Find distance from point to line in 3D

Take

AP⃗=P−A=(0+3, 2−1, 3+4)=(3,1,7).\vec{AP}=P-A=(0+3,\,2-1,\,3+4)=(3,1,7).AP=P−A=(0+3,2−1,3+4)=(3,1,7).

Distance from point PPP to the line through AAA with direction d⃗\vec dd is

d=∥AP⃗×d⃗∥∥d⃗∥.d=\frac{\lVert \vec{AP}\times \vec d\rVert}{\lVert \vec d\rVert}.d=∥d∥∥AP×d∥​.

Compute the cross product:

AP⃗×d⃗=∣i^j^k^317523∣.\vec{AP}\times \vec d= \begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & 1 & 7\\ 5 & 2 & 3 \end{vmatrix}.AP×d=​i^35​j^​12​k^73​​.

So,

AP⃗×d⃗=i^(1⋅3−7⋅2)−j^(3⋅3−7⋅5)+k^(3⋅2−1⋅5)\vec{AP}\times \vec d =\hat i(1\cdot 3-7\cdot 2)-\hat j(3\cdot 3-7\cdot 5)+\hat k(3\cdot 2-1\cdot 5)AP×d=i^(1⋅3−7⋅2)−j^​(3⋅3−7⋅5)+k^(3⋅2−1⋅5) =(−11,26,1).=(-11,26,1).=(−11,26,1).

Hence,

∥AP⃗×d⃗∥=(−11)2+262+12=121+676+1=798.\lVert \vec{AP}\times \vec d\rVert=\sqrt{(-11)^2+26^2+1^2} =\sqrt{121+676+1}= \sqrt{798}.∥AP×d∥=(−11)2+262+12​=121+676+1​=798​.

Also,

∥d⃗∥=52+22+32=38.\lVert \vec d\rVert=\sqrt{5^2+2^2+3^2}=\sqrt{38}.∥d∥=52+22+32​=38​.

Therefore,

d=79838=79838=21.d=\frac{\sqrt{798}}{\sqrt{38}}=\sqrt{\frac{798}{38}}=\sqrt{21}.d=38​798​​=38798​​=21​.
  1. Compute the area

Now,

Area=12⋅5⋅21=5212.\text{Area} = \frac12\cdot 5\cdot \sqrt{21}=\frac{5\sqrt{21}}{2}.Area=21​⋅5⋅21​=2521​​.

So if the area is mn\dfrac{m}{n}nm​ in the sense

mn=5212,\frac{m}{n}=\frac{5\sqrt{21}}{2},nm​=2521​​,

then

2m−521 n=0.2m-5\sqrt{21}\,n=0.2m−521​n=0.
  1. Check options
  • A: 2m−521n=02m-5\sqrt{21}n=02m−521​n=0 ✅
  • B: m−521n=0m-5\sqrt{21}n=0m−521​n=0 ❌
  • C: 5m−212n=05m-21\sqrt{2}n=05m−212​n=0 ❌
  • D: 5m−221n=05m-2\sqrt{21}n=05m−221​n=0 ❌

Therefore, the correct option is A.

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