COMEDK · Maths · 34. Three Dimensional Geometry
The equation of the perpendicular drawn from the point \(A(6, 1, 3)\) to the line \(\dfrac{x-1}{2} = \dfrac{2-y}{-1} = \dfrac{z-3}{2}\) is \(\dfrac{x-6}{a} = \dfrac{y-1}{b} = \dfrac{z-3}{c}\). If \(a, b, c\) are the possible integers such that \(a < 0\), then the value of \(a - b + 5c\) is:
- A \(-17\)
- B \(11\)
- C \(0\)
- D \(5\)
Answer & Solution
Correct Answer
(D) \(5\)
Step-by-step Solution
Detailed explanation
The given line can be rewritten in standard form as: \(\dfrac{x-1}{2} = \dfrac{y-2}{1} = \dfrac{z-3}{2}\) Let the foot of the perpendicular from \(A(6, 1, 3)\) to the given line be \(P\). Any point on the line can be taken as \(P(2\lambda + 1, \lambda + 2, 2\lambda + 3)\). The…
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