JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let a line \(l\) pass through the origin and be perpendicular to the lines \(l_1: \overrightarrow{ r }=(\hat{ i }-11 \hat{ j }-7 \hat{ k })+\lambda(\hat{ i }+2 \hat{ j }+3 \hat{ k }), \lambda \in R\) and \(l_2: \overrightarrow{ r }=(-\hat{ i }+\hat{ k })+\mu(2 \hat{ i }+2 \hat{ j }+\hat{ k }), \mu \in R\). If \(P\) is the point of intersection of \(l\) and \(l_1\), and \(Q (\alpha\) \(, \beta, \gamma)\) is the foot of perpendicular from \(P\) on \(l_2\), then \(9(\alpha+\beta+\gamma)\) is equal to \(..........\).
- A \(4\)
- B \(5\)
- C \(3\)
- D \(2\)
Answer & Solution
Correct Answer
(B) \(5\)
Step-by-step Solution
Detailed explanation
\(\text { Let } \ell=(0 \hat{ i }+0 \hat{ j }+0 \hat{ k })+\gamma( a \hat{ i }+ b \hat{ j }+ ck )\) \(=\gamma( a \hat{ i }+ b \hat{ j }+ c \hat{ k })\)…
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