JEE Mains · Maths · STD 12 - 11. three dimension geometry
A perpendicular is drawn from a point on the line \(\frac{{x - 1}}{2} = \frac{{y + 1}}{{ - 1}} = \frac{z}{1}\) to the plane \(x + y + z = 3\) such that the foot of the perpendicular \(Q\) also lies on the plane \(x -y + z = 3.\) Then the co-ordinates of \(Q\) are
- A \((2, 0, 1)\)
- B \((-1, 0, 4)\)
- C \((1, 0, 2)\)
- D \((4, 0, -1)\)
Answer & Solution
Correct Answer
(A) \((2, 0, 1)\)
Step-by-step Solution
Detailed explanation
Let point \(P\) on the line is \((2 \lambda+1-\lambda-1, \lambda)\) foot of perpendicular \(Q\) is given by \(\frac{x-2 \lambda-1}{1}=\frac{y+\lambda+1}{1}=\frac{z-\lambda}{1}=\frac{-(2 \lambda-3)}{3}\) \(\because\) \(Q\) lies on \(x+y+z=3\) and \(x-y+z=3\) \(\Rightarrow x+z=3\)…
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