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 equation of the line is: \[ \frac{2x + 4}{6} = \frac{y + 1}{2} = \frac{-2z + 6}{-4} = t \] This represents a parametric form of the line. We can express \( x, y, z \) in terms of the parameter \( t \): \[ x = \frac{6t - 4}{2}, \quad y = 2t - 1, \quad z = \frac{4t + 6}{2} \] Let the coordinates of point \( Q \) be \( (x_1, y_1, z_1) \). To find the distance between \( P(1, 2, 3) \) and \( Q(x_1, y_1, z_1) \), we use the distance formula: \[ \sqrt{(x_1 - 1)^2 + (y_1 - 2)^2 + (z_1 - 3)^2} = \frac{\sqrt{5}}{2} \] Substitute the values of \( x_1, y_1, z_1 \) from the parametric equations: \[ \sqrt{\left(\frac{6t - 4}{2} - 1\right)^2 + (2t - 1 - 2)^2 + \left(\frac{4t + 6}{2} - 3\right)^2} = \frac{\sqrt{5}}{2} \] Simplify the equation and solve for \( t \). After solving, we find: \[ t = \frac{2}{3} \] Substitute \( t = \frac{2}{3} \) back into the parametric equations to find the coordinates of \( Q \): \[ x_1 = \frac{6 \times \frac{2}{3} - 4}{2} = \frac{2}{3}, \quad y_1 = 2 \times \frac{2}{3} - 1 = \frac{5}{3}, \quad z_1 = \frac{4 \times \frac{2}{3} + 6}{2} = 3 \] Thus, the coordinates of \( Q \) are \( \left(\frac{2}{3}, \frac{5}{3}, 3\right) \).