JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the plane \(P\) contain the line \(2 x+y-z-3=0=5 x-3 y+4 z+9\) and be parallel to the line \(\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}\). Then the distance of the point \(A (8,-1,-19)\) from the plane P measured parallel to the line \(\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}\) is equal to \(............\).
- A \(26\)
- B \(25\)
- C \(24\)
- D \(23\)
Answer & Solution
Correct Answer
(A) \(26\)
Step-by-step Solution
Detailed explanation
Plane \(P \equiv P _1+\lambda P _2=0\) \((2 x+y-z-3)+\lambda(5 x-3 y)+4 z+9)=0\) \((5 \lambda+2) x+(1-3 \lambda) y+(4 \lambda-1) z+9 \lambda-3=0\) \(\overrightarrow{ n } \cdot \overrightarrow{ b }=0 \text { where } \overrightarrow{ b }(2,4,5)\)…
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