JEE Mains · Maths · STD 12 - 11. three dimension geometry
The plane \(2 x-y+z=4\) intersects the line segment joining the points \(A ( a ,-2\), 4) and \(B (2, b ,-3)\) at the point \(C\) in the ratio \(2: 1\) and the distance of the point \(C\) from the origin is \(\sqrt{5}\). If \(ab <0\) and \(P\) is the point \(( a - b , b , 2 b - a )\) then \(CP ^2\) is equal to :
- A \(\frac{17}{3}\)
- B \(\frac{16}{3}\)
- C \(\frac{73}{3}\)
- D \(\frac{97}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{17}{3}\)
Step-by-step Solution
Detailed explanation
\(A ( a ,-2,4), B (2, b ,-3)\) \(AC : CB =2: 1\) \(\Rightarrow C \equiv\left(\frac{ a +4}{3}, \frac{2 b -2}{3}, \frac{-2}{3}\right)\) \(C\) lies on \(2 x-y+2=4\) \(\Rightarrow \frac{2 a+8}{3}-\frac{2 b-2}{3}-\frac{2}{3}=4\) \(\Rightarrow a-b=2 \ldots(1)\)…
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