JEE Mains · Maths · STD 12 - 11. three dimension geometry
The distance of the point \((-1,9,-16)\) from the plane \(2 x+3 y-z=5\) measured parallel to the line \(\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12}\) is \(......\)
- A \(13 \sqrt{2}\)
- B \(31\)
- C \(26\)
- D \(20 \sqrt{2}\)
Answer & Solution
Correct Answer
(C) \(26\)
Step-by-step Solution
Detailed explanation
Equation of line \(\frac{x+1}{3}=\frac{y-9}{-4}=\frac{z+16}{12}\) \(G.P\) on line \((3 \lambda-1,-4 \lambda+9,12 \lambda-16)\) point of intersection of line \& plane \(6 \lambda-2-12 \lambda+27-12 \lambda+16=5\) \(\lambda=2\) Point \((5,1,8)\) Distance \(=\sqrt{36+64+576}=26\)
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