Question

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

Hint:

To find a point on a line at a specified distance from another point, parametrize the line and use the distance formula to find the parameter \( t \) corresponding to the given distance.

Answer 1

The equation of the line is given by: \[ \frac{x-1}{3} = \frac{y+1}{2} = \frac{z-4}{3} \] Let the common parameter be \( t \). Then, parametrize the coordinates of the point on the line: \[ x = 3t + 1, \quad y = 2t - 1, \quad z = 3t + 4 \] The distance between the point \( (-1, -1, 2) \) and a point on the line \( (3t + 1, 2t - 1, 3t + 4) \) is given by the formula: \[ d = \sqrt{(3t + 1 + 1)^2 + (2t - 1 + 1)^2 + (3t + 4 - 2)^2} \] Setting this equal to \( \sqrt{2} \), we solve for \( t \) to find the coordinates of the required point on the line.