JEE Mains · Maths · STD 12 - 11. three dimension geometry
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?
- A \(\alpha+2 \gamma=24\)
- B \(2 \alpha+\gamma=7\)
- C \(2 \alpha-\gamma=9\)
- D \(\alpha-2 \gamma=19\)
Answer & Solution
Correct Answer
(A) \(\alpha+2 \gamma=24\)
Step-by-step Solution
Detailed explanation
\(\overline{ b _1} \times \overline{ b _2}=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ -2 & 0 & 1 \\ 1 & 1 & -1\end{array}\right|=-\hat{ i }-\hat{ j }-2 \hat{ k }\)…
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