JEE Mains · Maths · STD 12 - 11. three dimension geometry
The distance of the point \((-1,2,3)\) from the plane \(\vec{r} .(\hat{i}-2 \hat{j}+3 \hat{k})=10\) parallel to the line of the shortest distance between the lines \(\overrightarrow{ r }=(\hat{ i }-\hat{ j })+\lambda(2 \hat{i}+\hat{ k })\) and \(\overrightarrow{ r }=(2 \hat{ i }-\hat{ j })+\mu(\hat{i}-\hat{j}+\hat{ k })\) is :
- A \(3 \sqrt{6}\)
- B \(3 \sqrt{5}\)
- C \(2 \sqrt{6}\)
- D \(2 \sqrt{5}\)
Answer & Solution
Correct Answer
(C) \(2 \sqrt{6}\)
Step-by-step Solution
Detailed explanation
Let \(L _1: \overrightarrow{ r }=(\hat{ i }-\hat{ j })+\lambda(2 \hat{ i }+\hat{ k })\) \(L_2: \overrightarrow{ r }=(2 \hat{ i }-\hat{ j })+\mu(\hat{ i }-\hat{ j }+\hat{ k })\)…
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