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 line is represented by the equation: \[ \frac{x-1}{3} = \frac{y+1}{2} = \frac{z-4}{3} \]. Introducing a common parameter \( t \), the coordinates of any point on the line can be parametrized as: \[ x = 3t + 1, \quad y = 2t - 1, \quad z = 3t + 4 \]. The distance between the fixed point \( (-1, -1, 2) \) and a general point on the line \( (3t + 1, 2t - 1, 3t + 4) \) is calculated using the distance formula: \[ d = \sqrt{(3t + 1 - (-1))^2 + (2t - 1 - (-1))^2 + (3t + 4 - 2)^2} \]. To find the specific point on the line, we set this distance equal to \( \sqrt{2} \) and solve for \( t \).