The question asks:
Find the line through $(3,1,-2)$ that intersects and is perpendicular to
$$x = -1 + t, y = -2 + t, z = -1 + t.$$
My thoughts:
Say the point of intersection is $(x_0,y_0,z_0)$, then my line can be of the form
$$L(s) = (3,1,-2) + (x_0- 3,y_0- 1,z_0+ 2)s$$
Then I tried setting up a system of equations at the intersection like this:
$$-1 + t = 3 + (x_0- 3)s$$
$$-2 + t = 1 + (y_0- 1)s$$
$$-1 + t = -2 + (z_0+ 2)s$$
And tried finding the point $(x_0,y_0,z_0)$, but I feel that I'm not on the right track. Could someone explain to me how to appropriately tackle this problem?
Best Answer
Here is an approach. Find a normal vector to the directional vector $(-1,-2,-1)$ of the given line, then use it (together with finding a point on the line $L$) to find the equation of the perpendicular line.