Question

(a) Find the point Q on the line \( \frac{2x + 4}{6} = \frac{y + 1}{2} = \frac{-2z + 6}{-4} \) at a distance of \( \frac{\sqrt{5}}{2} \) from the point \( P(1, 2, 3) \).

Hint:

Use the parametric form of the line to express the coordinates of the points and apply the distance formula. Solve for the parameter to find the exact coordinates of the point on the line.

Answer 1

The given line is in parametric form: \[ \frac{2x + 4}{6} = \frac{y + 1}{2} = \frac{-2z + 6}{-4} = t \] From this, we derive the parametric equations for \( x, y, z \): \[ x = \frac{6t - 4}{2}, \quad y = 2t - 1, \quad z = \frac{4t + 6}{2} \] Let point \( Q \) have coordinates \( (x_1, y_1, z_1) \). The distance between \( P(1, 2, 3) \) and \( Q(x_1, y_1, z_1) \) is given by: \[ \sqrt{(x_1 - 1)^2 + (y_1 - 2)^2 + (z_1 - 3)^2} = \frac{\sqrt{5}}{2} \] Substituting the parametric expressions for \( x_1, y_1, z_1 \): \[ \sqrt{\left(\frac{6t - 4}{2} - 1\right)^2 + (2t - 1 - 2)^2 + \left(\frac{4t + 6}{2} - 3\right)^2} = \frac{\sqrt{5}}{2} \] Solving this equation for \( t \) yields: \[ t = \frac{2}{3} \] Substituting \( t = \frac{2}{3} \) back into the parametric equations for \( x_1, y_1, z_1 \): \[ x_1 = \frac{6 \times \frac{2}{3} - 4}{2} = \frac{2}{3}, \quad y_1 = 2 \times \frac{2}{3} - 1 = \frac{5}{3}, \quad z_1 = \frac{4 \times \frac{2}{3} + 6}{2} = 3 \] Therefore, the coordinates of point \( Q \) are \( \left(\frac{2}{3}, \frac{5}{3}, 3\right) \).