# Find parametric equations for the line through the point that is perpendicular to the

by jcook735
Tags: equations, line, parametric, perpendicular, point
 P: 33 Find parametric equations for the line through the point that is perpendicular to the given line, and intersects this line. We are given the point (0, 2, 2) and the line x = 1 + t, y = 2 - t, z = 2t. We are also given one of the parametric equations, that being x=3s. I am looking for the y and z parametric equations. I figured that the equation for the first line is = (1, 2, 0) + t(1, -1, 2) and the equation for the second line is (0,2,2) + s(3,b,c). Beyond that, I have no idea what to do.
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P: 6,992
 Quote by jcook735 Find parametric equations for the line through the point that is perpendicular to the given line, and intersects this line. We are given the point (0, 2, 2) and the line x = 1 + t, y = 2 - t, z = 2t. We are also given one of the parametric equations, that being x=3s. I am looking for the y and z parametric equations. I figured that the equation for the first line is = (1, 2, 0) + t(1, -1, 2) and the equation for the second line is (0,2,2) + s(3,b,c). Beyond that, I have no idea what to do.
Have you had how to calculate the vector component of one vector on another? If so, here's a hint:

Consider a point P on the given line, for example P = (1,2,0). Consider the vector V from that point to your point (0,2,2). Calculate the vector component of V on your line. You can use that and your point P to calculate the point Q where the perpendicular line would intersect. Knowing P and Q the rest is easy...
 P: 33 I feel like I am missing something really obvious, but I don't understand how i could use point P and the vector component V to find point Q. The vector component V is (-1,0,2), correct?
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Thanks
PF Gold
P: 6,992

## Find parametric equations for the line through the point that is perpendicular to the

 Quote by LCKurtz Have you had how to calculate the vector component of one vector on another? If so, here's a hint: Consider a point P on the given line, for example P = (1,2,0). Consider the vector V from that point to your point (0,2,2). Calculate the vector component of V on your line. You can use that and your point P to calculate the point Q where the perpendicular line would intersect. Knowing P and Q the rest is easy...
 Quote by jcook735 I feel like I am missing something really obvious, but I don't understand how i could use point P and the vector component V to find point Q. The vector component V is (-1,0,2), correct?
"Vector component V" doesn't make sense.

It is the displacement vector V itself that is <-1,0,2>. That isn't the same thing as the vector projection of V on your line.

Call your point (0,2,2) S. Draw a picture of a line noting the point P(1,2,0) on the line and the point S(0,2,2) off the line and the displacement vector V going from P to S. If you drop a perpendicular from S to the line, hitting it at Q, you get a little right triangle. Note that no matter what point on the line you use for P, you always get the same Q.

Also notice that if you add the components of the vector PQ to the coordinates of P you will get the coordinates of Q. And once you have Q you can write the equation of the line you seek.

The vector PQ is called the vector projection of V on the line. It is what you need to calculate to solve your problem, which is why I asked if you had studied vector projections yet. Does that help?
 P: 33 Ohhhhh yes I understand now. Thank you!!

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