Let P = (1, 1, -1), Q = (2, -2, -3), and R = (-1, 0, 4) be the vertices of a triangle.
Let L be a line passing through P and Q.
Find:
The point on the line x = y + 1 = z – 2 closest to the line L.
[Math] Find the closest point on one line to another line in 3D
calculus
Related Question
- [Math] Find the closest point to the origin
- [Math] Find the equation of the line passing through the point (3,4) which cuts the first quarter a triangle with minimum area
- [Math] Find the point on the line $x + 4y − 7 = 0$ which is closest to the point $(-2, -2)$
- [Math] Find Parametric Equations for a line passing through point and intersecting line at 90 degrees
- [Math] Finding the point on a line closest to another point not on the line
- [Math] Point On a Plane Closest to a Point
Best Answer
Hint: A parameterized version of L is $(1+t,1-3t,-1-2t)$. The other line is $(s, 1+s, s-2)$. So you can just calculate the squared distance, take the derivative with respect to s and t, and get two equations in two unknowns.