JEE Mains · Maths · STD 12 - 11. three dimension geometry
The point of intersection \(C\) of the plane \(8 x+y+2 z=0\) and the line joining the points \(A (-3,-6,1)\) and \(B (2,4,-3)\) divides the line segment \(AB\) internally in the ratio \(k : 1\). If \(a , b , c\) \((| a |,| b |,| c |\) are coprime) are the direction ratios of the perpendicular from the point \(C\) on the line \(\frac{1- x }{1}=\frac{ y +4}{2}=\frac{z+2}{3}\), then \(|a + b + c|\) is equal to \(.............\).
- A \(100\)
- B \(10\)
- C \(1000\)
- D \(200\)
Answer & Solution
Correct Answer
(B) \(10\)
Step-by-step Solution
Detailed explanation
Plane: \(8 x+y+2 z=0\) Given line \(AB : \frac{ x -2}{5}=\frac{ y -4}{10}=\frac{ z +3}{-4}=\lambda\) Any point on line \((5 \lambda+2,10 \lambda+4,-4 \lambda-3)\) Point of intersection of line and plane \(8(5 \lambda+2)+10 \lambda+4-8 \lambda-6=0\) \(\lambda=-\frac{1}{3}\)…
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