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STD 12 - 11. three dimension geometry — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 11. three dimension geometry1 Mark
MCQ
Let the shortest distance between the lines $L : \frac{ x -5}{-2}=\frac{ y -\lambda}{0}=\frac{ z +\lambda}{1}, \lambda \geq 0$ and $L _1: x +1= y -$ $1=4-z$ be $2 \sqrt{6}$. If $(\alpha, \beta, \gamma)$ lies on $L$, then which of the following is NOT possible?
  • ✓
    $\alpha+2 \gamma=24$
  • B
    $2 \alpha+\gamma=7$
  • C
    $2 \alpha-\gamma=9$
  • D
    $\alpha-2 \gamma=19$
✓

Answer

Correct option: A.
$\alpha+2 \gamma=24$
a
$\overline{ b _1} \times \overline{ b _2}=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ -2 & 0 & 1 \\ 1 & 1 & -1\end{array}\right|=-\hat{ i }-\hat{ j }-2 \hat{ k }$

$\begin{array}{l}\overline{ a _2}-\overline{ a _1}=6 \hat{ i }+(\lambda-1) \hat{ j }+(-\lambda-4) \hat{ k } \\ 2 \sqrt{6}=\left|\frac{-6-\lambda+1+2 \lambda+8}{\sqrt{1+1+4}}\right| \\ |\lambda+3|=12 \Rightarrow \lambda=9,-15 \\ \alpha=-2 k +5, \gamma= k -\lambda \text { where } k \in R \\ \Rightarrow \alpha+2 \gamma=5-2 \lambda=-13,35\end{array}$

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