Equation of a line through a point and another line at 90 degrees.

• IcedCore

Homework Statement

Find the equation of the line through the point (-5,-4,2) that intersects the line at r = (7,-13,8) +t(1,2,-2) at 90 degrees.

Homework Equations

I guess that using the dot product for the direction vector of the line needed, (x,y,z) , with the direction of the given line will = 0. Then i get lost as to what I should be doing afterwards, in incorporating the point given with the variables that result from the dot product.

The Attempt at a Solution

So i did (x,y,z) dot (1,2,-2) to get x +2y -2z = 0
at this point, I'm not sure where to do.

Btw, the answer is r = (-5,-4,2) +t(14,-5,2).

Homework Statement

Find the equation of the line through the point (-5,-4,2) that intersects the line at r = (7,-13,8) +t(1,2,-2) at 90 degrees.

Homework Equations

I guess that using the dot product for the direction vector of the line needed, (x,y,z) , with the direction of the given line will = 0. Then i get lost as to what I should be doing afterwards, in incorporating the point given with the variables that result from the dot product.

The Attempt at a Solution

So i did (x,y,z) dot (1,2,-2) to get x +2y -2z = 0
at this point, I'm not sure where to do.

Btw, the answer is r = (-5,-4,2) +t(14,-5,2).

Your whole problem will be solved if you can figure out the direction from Q that is perpendicular to your line. Call P = (7,-13,8) and Q = (-5,-4,2) and V the vector from P to Q. Call the direction vector of your line D= <1,2,-2>. You know
W= VxD is perpendicular to both V and D.

Now think about what direction WxD would be.

I would do this in a slightly different way. The given line has "direction vector" <1, 2, -2>. A line that passes throught (-5, -4, 2) and is perpendicular to the given line must lie in the plane having that vector as normal vector and containing that point: that plane is 1(x+ 5)+ 2(y+ 4)- 2(z- 2)= 0. Find the point at which the given line intersects that plane and use that and (-5, -5, 2) to determine the line.

I tried both ways, but I'm getting slightly different answers. Could you please show me some clear steps because I'm getting somewhat confused.

A note to take into consideration is that we haven't really touched upon 3D intersections, using perpendicular vectors. Never.

I tried both ways, but I'm getting slightly different answers. Could you please show me some clear steps because I'm getting somewhat confused.

If you tried my suggestion and made no arithmetic errors, you can shorten your final vector by dividing it by 9 to get your suggested direction. Was that your problem?

V = Q-P = <-12,9,-6>
W = V x D = <-6,-30,-33>
Normal = W x D = <126,-45,18>

Divide by 9 to shorten it: <14,-5,2>. The numbers would likely be smaller if I had divided V by 3 to begin with.

Ahh, thank you so much. I realized that I made a mistake half way through, but i changed the wrong number in the last cross. Thanks! By the way, if you don't mind, why did you choose to complete the operations that you did. My teacher only gives very basic explanations, if any, and sometimes I do get confused.

Ahh, thank you so much. I realized that I made a mistake half way through, but i changed the wrong number in the last cross. Thanks! By the way, if you don't mind, why did you choose to complete the operations that you did. My teacher only gives very basic explanations, if any, and sometimes I do get confused.

There are several ways to work that problem. You could have found the vector projection of V = PQ on your line and subtracted that from V to get the direction. You could use the method HallsOfIvy suggested. Or you could use what I suggested. They all work.

The reason the method I showed you works is that W = V x D is perpendicular to the plane of V and D, so when you cross W with D, the result, being perpendicular to W must be back parallel to the plane of V and D. And since it is perpendicular to D, it is the required direction. It might be helpful for you to draw a picture.