JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the image of the point \(P(1, 6, a)\) in the line \(L: \dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-a+1}{b}\), \(b > 0\), be \(\left(\dfrac{a}{3}, 0, a+c\right)\). If \(S(\alpha, \beta, \gamma)\), \(\alpha > 0\), is the point on \(L\) such that the distance of \(S\) from the foot of perpendicular from the point \(P\) on \(L\) is \(2\sqrt{14}\), then \(\alpha + \beta + \gamma\) is equal to:
- A \(19\)
- B \(20\)
- C \(21\)
- D \(22\)
Answer & Solution
Correct Answer
(C) \(21\)
Step-by-step Solution
Detailed explanation
Let the foot of the perpendicular from \(P(1, 6, a)\) to the line \(L\) be \(M\). Since the image of \(P\) is \(P'\left(\dfrac{a}{3}, 0, a+c\right)\), \(M\) is the midpoint of \(P\) and \(P'\).…
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