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's equation is \( \frac{x-1}{3} = \frac{y+1}{2} = \frac{z-4}{3} \). Let \( t \) be the common parameter. The parametric equations for a point on the line are \( x = 3t + 1, y = 2t - 1, z = 3t + 4 \). The distance \( d \) between \( (-1, -1, 2) \) and a point on the line \( (3t + 1, 2t - 1, 3t + 4) \) is \( d = \sqrt{(3t + 1 + 1)^2 + (2t - 1 + 1)^2 + (3t + 4 - 2)^2} \). Setting \( d = \sqrt{2} \) allows us to solve for \( t \) and determine the coordinates of the desired point on the line.