ExamBro
ExamBro
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.

  1. A \(\left(-\dfrac{14}{9} , -\dfrac{28}{9} , -\dfrac{28}{9}\right)\)
  2. B \((6 , 1 , 5)\)
  3. C \((2 , 4 , 4)\)
  4. D \((1 , 2 , 2)\)
Verified Solution

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 \(L_1\).…