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3D Geometry question

2025 · 28 Jan · Shift 2 · Q30
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  5. /2025 · 28 Jan · Shift 2 · Q30

3D Geometry question

2025 · 28 Jan · Shift 2 · Q30

JEE MainMathematics3D GeometryMCQ+4 / −1
The square of the distance of the point (157,327,7)\left( \frac{15}{7}, \frac{32}{7}, 7 \right)(715​,732​,7) from the line x+13=y+35=z+57\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}3x+1​=5y+3​=7z+5​ in the direction of the vector i^+4j^+7k^\hat{i} + 4\hat{j} + 7\hat{k}i^+4j^​+7k^ is:
  1. A
    66
  2. B
    54
  3. C
    41
  4. D
    44
View written solutionFree

Correct answer: A

  1. Interpret the line and given direction

The line is

x+13=y+35=z+57=t.\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}=t.3x+1​=5y+3​=7z+5​=t.

So a general point on the line is

(−1+3t, −3+5t, −5+7t).(-1+3t,\,-3+5t,\,-5+7t).(−1+3t,−3+5t,−5+7t).

Its direction vector is

d⃗=(3,5,7).\vec d=(3,5,7).d=(3,5,7).

The given point is

P(157,327,7),P\left(\frac{15}{7},\frac{32}{7},7\right),P(715​,732​,7),

and the given direction is

v⃗=(1,4,7).\vec v=(1,4,7).v=(1,4,7).

We need the distance of the point from the line measured along the direction v⃗\vec vv. That means: draw through PPP a line parallel to v⃗\vec vv, and find where it meets the given line. The required distance is the length from PPP to that intersection point.

  1. Equation of the line through PPP parallel to v⃗\vec vv

A point on this line is

(157+λ,  327+4λ,  7+7λ).\left(\frac{15}{7}+\lambda,\;\frac{32}{7}+4\lambda,\;7+7\lambda\right).(715​+λ,732​+4λ,7+7λ).

This must intersect the given line, so for some parameters ttt and λ\lambdaλ,

157+λ=−1+3t...(1)\frac{15}{7}+\lambda=-1+3t \quad ...(1)715​+λ=−1+3t...(1) 327+4λ=−3+5t...(2)\frac{32}{7}+4\lambda=-3+5t \quad ...(2)732​+4λ=−3+5t...(2) 7+7λ=−5+7t...(3)7+7\lambda=-5+7t \quad ...(3)7+7λ=−5+7t...(3)
  1. Solve for λ\lambdaλ and ttt

From (3):

7+7λ=−5+7t7+7\lambda=-5+7t7+7λ=−5+7t 12+7λ=7t12+7\lambda=7t12+7λ=7t t=127+λ.t=\frac{12}{7}+\lambda.t=712​+λ.

Substitute into (1):

157+λ=−1+3(127+λ)\frac{15}{7}+\lambda=-1+3\left(\frac{12}{7}+\lambda\right)715​+λ=−1+3(712​+λ) 157+λ=−1+367+3λ\frac{15}{7}+\lambda=-1+\frac{36}{7}+3\lambda715​+λ=−1+736​+3λ 157+λ=297+3λ\frac{15}{7}+\lambda=\frac{29}{7}+3\lambda715​+λ=729​+3λ −2=2λ-2=2\lambda−2=2λ λ=−1.\lambda=-1.λ=−1.

Then

t=127−1=57.t=\frac{12}{7}-1=\frac{5}{7}.t=712​−1=75​.

Check in (2):

327+4(−1)=327−287=47,\frac{32}{7}+4(-1)=\frac{32}{7}-\frac{28}{7}=\frac{4}{7},732​+4(−1)=732​−728​=74​,

while

−3+5⋅57=−217+257=47.-3+5\cdot\frac{5}{7}=-\frac{21}{7}+\frac{25}{7}=\frac{4}{7}.−3+5⋅75​=−721​+725​=74​.

So it is consistent.

  1. Find the required distance

Since λ=−1\lambda=-1λ=−1, the intersection point lies on the line through PPP in direction v⃗\vec vv at parameter −1-1−1. Hence the displacement from PPP to the line is

λv⃗=−1(1,4,7)=(−1,−4,−7).\lambda\vec v=-1(1,4,7)=(-1,-4,-7).λv=−1(1,4,7)=(−1,−4,−7).

So the distance is

∣λ∣ ∣v⃗∣=1⋅12+42+72=66.|\lambda|\,|\vec v|=1\cdot\sqrt{1^2+4^2+7^2}=\sqrt{66}.∣λ∣∣v∣=1⋅12+42+72​=66​.

Therefore, the square of the distance is

66.66.66.
  1. Option check
  • A: 666666 ✅
  • B: 545454 ❌
  • C: 414141 ❌
  • D: 444444 ❌

So the correct option is A.

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