JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let a plane \(P\) contain two lines \(\overrightarrow{ r }=\hat{ i }+\lambda(\hat{ i }+\hat{ j }), \lambda \in R\) and \(\overrightarrow{ r }=-\hat{j}+\mu(\hat{j}-\hat{ k }), \mu \in R\) If \(Q (\alpha, \beta, \gamma)\) is the foot of the perpendicular drawn from the point \(M (1,0,1)\) to \(P ,\) then \(3(\alpha+\beta+\gamma)\) equals
- A \(6\)
- B \(8\)
- C \(5\)
- D \(10\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
Dr's normal to plane \(=\left|\begin{array}{ccc}i & j & k \\1 & 1 & 0 \\0 & 1 & -1\end{array}\right|=-\hat{i}+\hat{j}+\hat{k}\) Equation of plane \(-1(x-1)+1(y-0)+1(z-0)=0\) \(x-y-z-1=0\) Now \(\frac{\alpha-1}{1}=\frac{\beta-0}{-1}=\frac{\gamma-1}{-1}=-\frac{(1-0-1-1)}{3}\)…
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