JEE Mains · Maths · STD 12 - 11. three dimension geometry
The line of shortest distance between the lines \(\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}\) and \(\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}\) makes an angle of \(\cos ^{-1}\left(\sqrt{\frac{2}{27}}\right)\) with the plane \(P: a x-y-\) \(z=0\), \((a>0)\). If the image of the point \((1,1,-5)\) in the plane \(P\) is \((\alpha, \beta, \gamma)\), then \(\alpha+\beta-\gamma\) is equal to \(........\)
- A \(4\)
- B \(5\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\(DR's\) of line of shortest distance \(\left|\begin{array}{lll}\hat{ i } & \hat{ j } & \hat{ k } \\0 & 1 & 1 \\2 & 2 & 1\end{array}\right|=-\hat{ i }+2 \hat{ j }-2 \hat{ k }\)angle between line and plane is \(\cos ^{-1} \sqrt{\frac{2}{27}}=\alpha\)…
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