COMEDK · Maths · 34. Three Dimensional Geometry
Let P be a point on the line \(L_1: \dfrac{x-2}{2} = y + 1 = \dfrac{z-1}{2}\) such that its distance from the point \(A(2 , -1 , 1)\) is 6 units.
Given that \(x\)-coordinate of P is greater than 2,
Find the coordinates of point Q on the line \(L_2: x - 1 = \dfrac{y-2}{2} = \dfrac{z-2}{2}\) such that Q is the closest point to P.
- A \(\left(-\dfrac{14}{9} , -\dfrac{28}{9} , -\dfrac{28}{9}\right)\)
- B \((6 , 1 , 5)\)
- C \((2 , 4 , 4)\)
- D \((1 , 2 , 2)\)
Answer & Solution
Correct Answer
(C) \((2 , 4 , 4)\)
Step-by-step Solution
Detailed explanation
The line \(L_1\) is given by \(\dfrac{x-2}{2} = \dfrac{y+1}{1} = \dfrac{z-1}{2}\). The direction ratios of \(L_1\) are \((2, 1, 2)\), so its direction cosines are \(\left(\dfrac{2}{3}, \dfrac{1}{3}, \dfrac{2}{3}\right)\). The point \(A\) is \((2, -1, 1)\), which lies on \(L_1\).…
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