Perpendicular equation to two lines

[GreenFigure]

So I've been given this task at the university. I need to find perpendicular canonical equations common for both lines. Can't figure out how to do it, can anyone help me?

Line1:
{3x+2y-4z+1=0
{5x-y+2z-4=0
Line2:
{2x-3y+z+8=0
{4x-2y+5=0

 

[GreenFigure]

Well

Well what im asking is, I've got two lines, those are the equations, and I need to find a common canonical equation of a line that is perpendicular to both of those lines. Sorry for my bad english
and the "{" means equation system,

don't really know the terms for it in english and how do I show it to you. I've been trying to do this for few hours already and my head is all messed up.

\(\displaystyle \begin{cases}3x\, +\, 2y\, -\, 4z\, -\, 1\, =\, 0\\5x\, -\, y\, +\, 2z\, -\, 4\, =\, 0\end{cases}\)

Here's an attachment picture how it looks like.

 

[HallsofIvy]

Line1:
{3x+2y-4z+1=0
{5x-y+2z-4=0
Line2:
{2x-3y+z+8=0
{4x-2y+5=0

Your "canonical equations" are the equations of two planes, the line being the line of intersection. Because I am more comfortable with parametric equations the first thing I would do is change to parametric equations!

If we multiply the second equation for "Line 1" by 2 and add to the first, both y and z are eliminated:
(3x+ 2y- 4x+ 1)+ (10x- 2y+ 4z- 8)= 13x- 7= 0 so x must be the constant 7/13. Putting that into the second equation, 35/13- y+ 2z- 4= 0 so y= 17/13- 2z. we can write the line as x= 7/13, y= 17/13- 2t, z= t where t is the parameter. That line has direction vector \(\displaystyle -2\vec{j}+ \vec{k}\).

The second equation of "Line 2" has no "z" so we can immediately write y= 2x+ 5/2. Putting that into the first equation, 2x- 3(2x+ 5/2)+ z+ 8= -4x- 15/2+ z+ 8= 0 so z= 4x- 1/2. We can write x= t, y= 2t+ 5/2, z= 4t- 1/2 as parametric equations and that has direction vector \(\displaystyle \vec{i}+ 2\vec{j}+ 4\vec{k}\).

The vector perpendicular to both lines is the cross product of those two vectors. Now, you need to find a, b, c such that x= pt+ a, y= qt+ b, z= rt+ c (with p, q, t from the cross product) so that it intersects both lines.

 

[GreenFigure]

Thanks Ivy, while someone else might post any more tips ill try that out! thank you

 

[stapel]

Thanks Ivy, while someone else might post any more tips ill try that out! thank you

If you're not familiar with parametrization or vectors, you must have been given some other method for this sort of exercise. Please reply with that information. Thank you!

 

[GreenFigure]

If you're not familiar with parametrization or vectors, you must have been given some other method for this sort of exercise. Please reply with that information. Thank you!

well we are familiar but we haven't been given any information how to find perpendicular equations, thats the problem. Even my friend from second study course was like - woah, I dont have these written down. We just were introduced to it, some things were proven, few theorems and thats all, nothing was said about perpendicular... well there was an equation about a line that goes through A1(x1,y1,z1) and is perpendicular to a plane ax+bu+cz+d=0, and the equation is (x-x1)/a=(y-y1)b=(z-z1)/c tho even if its possible idk how to use it