# Please help me solve this problem. I have been stuck with it for two days now!

• Sep 24th 2012, 10:43 PM
andur92
Please help me solve this problem. I have been stuck with it for two days now!
Find the vector and Cartesian equation of the line passing through the point (1, 0,5) and is perpendicular to the line (x+2)/3 = (y-2)/4 = (z+3)/5 .

Also how wold solve these two equations:

l+2m+ 3n = 0
-3l+2m+5n = 0

The answer in my book is given as

l/10-6 = m/-9-5 = n/2+6

Here l,m,n are direction cosines.
• Sep 24th 2012, 11:05 PM
Siron
Re: Please help me solve this problem. I have been stuck with it for two days now!
To make it easier to represent the problem you can draw a sketch of it. A cartesian line is determined by one point and two direction vectors or one point and his normal vector (perpendicular to the plane). Which option do you think we can use here and why?
• Sep 26th 2012, 07:41 AM
HallsofIvy
Re: Please help me solve this problem. I have been stuck with it for two days now!
You are given a line in three dimensions in "symmetric form", (x+2)/3 = (y-2)/4 = (z+3)/5. You can write that as \$\displaystyle (x+ 2)/3= t\$, \$\displaystyle (y-2)/4= t\$, and \$\displaystyle (z+3)/5= t\$ which is the same as the parametric equations \$\displaystyle x= 3t- 2\$, [tex], \$\displaystyle y= 4t+ 2\$, and \$\displaystyle z= 5t- 3\$. You should know from that, as well as from the symmetric equations, that <3, 4, 5>, the coefficients of t in the parametric equations, and the denominators in the symmetric form, is a vector in the direction of the line.

Any plane perpendicular to that line is of the form 3x+ 4y+ 5z= C for some constant C. The perpendicular to the given line must lie in such a plane so the point (1, 0, 5), on the line, is on the plane and we must have 3(1)+ 4(0)+ 5(5)= 3+ 25= 28= C. That is, the line perpendicular to the given line must lie on the plane 3x+ 4y+ 5z= 28. Find the point where the given line intersects that plane and it, together with (1, 0, 5), give you two points that determine the line.