JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the distance of the point \((a, 2, 5)\) from the image of the point \((1, 2, 7)\) in the line \(\dfrac{x}{1} = \dfrac{y-1}{1} = \dfrac{z-2}{2}\) is \(4\), then the sum of all possible values of \(a\) is equal to :
- A \(11\)
- B \(9\)
- C \(6\)
- D \(4\)
Answer & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
Let \(P(1, 2, 7)\) be the given point and the given line be \(L: \dfrac{x}{1} = \dfrac{y-1}{1} = \dfrac{z-2}{2} = \lambda\). Any point on the line can be written as \(F(\lambda, \lambda+1, 2\lambda+2)\). If \(F\) is the foot of the perpendicular from \(P\) to the line, the…
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