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QJEEM24Ma081f: JEE Main 2024 Mathematics - Class 12 JEEM24Ma081f Solution: Consider the line passing through the points and . The distance …

Consider the line L\mathrm{L} passing through the points (1,2,3)(1,2,3) and (2,3,5)(2,3,5) . The distance of the point (113,113,193)\left(\frac{11}{3}, \frac{11}{3}, \frac{19}{3}\right) from the line L\mathrm{L} along the line 3x−112=3y−111=3z−192\frac{3 x-11}{2}=\frac{3 y-11}{1}=\frac{3 z-19}{2} is equal to

Multiple Choice Options

  • A 6
  • B 3
  • C 5
  • D 4

Step-by-Step Verified Solution

L:x−11=y−21=z−32=μL: \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{2}=\mu

Measured along L2:x−11323=y−11313=z−19323=λL_2: \frac{x-\frac{11}{3}}{\frac{2}{3}}=\frac{y-\frac{11}{3}}{\frac{1}{3}}=\frac{z-\frac{19}{3}}{\frac{2}{3}}=\lambda

Any point on L1:(μ+1,μ+2,2μ+3)L_1:(\mu+1, \mu+2,2 \mu+3)

Any point on L2(23λ+113,λ3+113,23λ+193)L_2\left(\frac{2}{3} \lambda+\frac{11}{3}, \frac{\lambda}{3}+\frac{11}{3}, \frac{2}{3} \lambda+\frac{19}{3}\right)

JEE Main 2024 (Online) 9th April Evening Shift Mathematics - 3D Geometry Question 36 English Explanation

Now

amp;μ+1=23λ+113amp;μ+2=λ3+113λ=−3amp;μ=23\begin{aligned}& \mu+1=\frac{2}{3} \lambda+\frac{11}{3} \\& \frac{\mu+2=\frac{\lambda}{3}+\frac{11}{3}}{\lambda=-3} \\& \mu=\frac{2}{3} \\\end{aligned}

amp; Point on L=(53,83,133)amp;d=(113−53)2+(83−113)2+(193−133)2amp;d=4+1+4amp;d=3\begin{aligned}& \text { Point on } L=\left(\frac{5}{3}, \frac{8}{3}, \frac{13}{3}\right) \\& d=\sqrt{\left(\frac{11}{3}-\frac{5}{3}\right)^2+\left(\frac{8}{3}-\frac{11}{3}\right)^2+\left(\frac{19}{3}-\frac{13}{3}\right)^2} \\& d=\sqrt{4+1+4} \\& d=3\end{aligned}

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