Register to reply 
Find parametric equations for the line through the point that is perpendicular to the 
Share this thread: 
#1
Sep710, 07:06 PM

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. 


#2
Sep710, 11:22 PM

HW Helper
Thanks
PF Gold
P: 7,568

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... 


#3
Sep810, 12:02 AM

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?



#4
Sep810, 11:48 AM

HW Helper
Thanks
PF Gold
P: 7,568

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


#5
Sep810, 06:10 PM

P: 33

Ohhhhh yes I understand now. Thank you!!



Register to reply 
Related Discussions  
Find parametric equations and symmetric equations for the line  Calculus & Beyond Homework  2  
Find parametric equations for the tangent line at the point  Calculus & Beyond Homework  4  
Line passes through a point parallet to parametric equations  Calculus & Beyond Homework  4  
Find Equation of Plane perpendicular to line passing through a point  Calculus & Beyond Homework  9  
Find Cartesian equation of plane containging line with parametric equations and point  Precalculus Mathematics Homework  3 