JEE Mains · Maths · STD 12 - 11. three dimension geometry
The distance of the point \(P(3,4,4)\) from the point of intersection of the line joining the points \(\mathrm{Q}(3,-4,-5)\) and \(\mathrm{R}(2,-3,1)\) and the plane \(2 \mathrm{x}+\mathrm{y}+\mathrm{z}=7\), is equal to \(.....\)
- A \(4\)
- B \(5\)
- C \(6\)
- D \(7\)
Answer & Solution
Correct Answer
(D) \(7\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{QR}}: \frac{x-3}{1}=\frac{y+4}{-1}=\frac{z+5}{-6}=r\) \(\Rightarrow(x, y, z) \equiv(r+3,-r-4,-6 r-5)\) Now, satisfying it in the given plane. \(2(r+3)+(-r-4)+(-6 r-5)=7\) We get \(r=-2\) so, required point of intersection is \(\mathrm{T}(1,-2,7) .\)…
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