JEE Advanced

Question 18 — JEE Main 21 Jan 2026 S Ii

Viewing 2 posts - 1 through 2 (of 2 total)
  • Author
    Posts
  • #11624
    shahkb4
    Keymaster

    Let the line $\mathrm{L}_{1}$ be parallel to the vector $-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ and pass through the point (2,6,7) and the line $\mathrm{L}_{2}$ be parallel to the vector $2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and pass through the point (4,3,5). If the line $\mathrm{L}_{3}$ is parallel to the vector $-3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+16 \mathrm{k}$ and intersects the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ at the points C and D , respectively, then $|\overrightarrow{\mathrm{CD}}|^{2}$ is equal to
    [JEE MAIN 21 Jan 2026 S II]
    (1) 171
    (2) 290
    (3) 312
    (4) 89

    #11625
    shahkb4
    Keymaster

    Solution

    line $\mathrm{L}_{1} \frac{\mathrm{x}-2}{-3}=\frac{\mathrm{y}-6}{2}=\frac{\mathrm{z}-7}{4}=\lambda_{1}$
    line $\mathrm{L}_{2} \frac{\mathrm{x}-4}{2}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}-5}{3}=\lambda_{2}$
    Point $\mathrm{C}\left(-3 \lambda_{1}+2,2 \lambda_{1}+6,4 \lambda_{1}+7\right)$ on line $\mathrm{L}_{1}$
    Point $\mathrm{D}\left(2 \lambda_{2}+4, \lambda_{2}+3,3 \lambda_{2}+5\right)$ on line $\mathrm{L}_{2}$

    \[ \begin{aligned} & \mathrm{CD} \xrightarrow{\mathrm{drs}} 2 \lambda_{2}+3 \lambda_{1}+2, \lambda_{2}-2 \lambda_{1}-3,3 \lambda_{2}-4 \lambda_{1}-2 \\ & \text { or }-3,5,16 \\ & \frac{2 \lambda_{2}+3 \lambda_{1}+2}{-3}=\frac{\lambda_{2}-2 \lambda_{1}-3}{5}=\frac{3 \lambda_{2}-4 \lambda_{1}-2}{16} \end{aligned} \]

    From first \& middle

    \[ \begin{aligned} & 10 \lambda_{2}+15 \lambda_{1}+10=-3 \lambda_{2}+6 \lambda_{1}+9 \\ & \Rightarrow 13 \lambda_{2}+9 \lambda_{1}+1=0 \end{aligned} \]

    From middle \& third

    \[ \begin{aligned} & 16 \lambda_{2}-32 \lambda_{1}-48=15 \lambda_{2}-20 \lambda_{1}-10 \\ & \Rightarrow \lambda_{2}=12 \lambda_{1}+38 \\ & 13\left(12 \lambda_{1}+38\right)+9 \lambda_{1}+1=0 \\ & \Rightarrow 165 \lambda_{1}+495=0 \Rightarrow \lambda_{1}=-3 \\ & \Rightarrow \lambda_{2}=-36+38 \Rightarrow \lambda_{2}=2 \text { Ans. } \\ & \lambda_{1}=-3, \lambda_{2}=2 \end{aligned} \]

    we get $\mathrm{C}(11,0,-5) \mathrm{D}(8,5,11)$

    \[ \mathrm{CD}^{2}=9+25+256=290 \text { Answer }(2) \]
Viewing 2 posts - 1 through 2 (of 2 total)
  • You must be logged in to reply to this topic.