JEE Mains · Maths · STD 12 - 11. three dimension geometry
The distance of the point \((1,1,9)\) from the point of intersection of the line \(\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}\) and the plane \(x+y+z=17\) is
- A \(2 \sqrt{19}\)
- B \(19 \sqrt{2}\)
- C \(38\)
- D \(\sqrt{38}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{38}\)
Step-by-step Solution
Detailed explanation
Let \(\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=t\) \(\Rightarrow x=3+t, y=2 t+4, z=2 t+5\) for point of intersection with \(x+y+z=17\) \(3+ t +2 t +4+2 t +5=17\) \(\Rightarrow \quad 5 t =5 \Rightarrow t =1\) \(\Rightarrow\) point of intersection is \((4,6,7)\) distance between…
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