Vector equation perpendicular to two equations

  • #1
52
0

Main Question or Discussion Point

How do I find the vector equation of the line which passes through (-3/2,-3/2,1/2) and is perpendicular to both x+1=y/3=-z and 2x+1=2y+1=z-5/2

I know how to do it using one equation but I am unsure about how to do it using two equations

Thanks
 

Answers and Replies

  • #2
2,788
586
Find tangent vector to those lines and take their cross product. Then find the equation of a line with a tangent equal to that cross product which passes from that point.
 
  • #3
52
0
Find tangent vector to those lines and take their cross product. Then find the equation of a line with a tangent equal to that cross product which passes from that point.
Could you please show an example I cant find anything about tangent vectors in my book

Thanks
 
  • #4
2,788
586
For example for the line ##x+1=\frac y 3=-z##: At first we should transform it to the parametric form, simply write ## z=-t ##, then we'll have ## \frac y 3=t \Rightarrow y=3t ## and ## x=t-1 ##. So each point on the line has coordinates of the form ## (t-1,3t,-t) ## and the tangent vector to this line is simply the derivative of the latter expression w.r.t. t which is ## (1,3,-1) ##.
 
  • #5
52
0
For example for the line ##x+1=\frac y 3=-z##: At first we should transform it to the parametric form, simply write ## z=-t ##, then we'll have ## \frac y 3=t \Rightarrow y=3t ## and ## x=t-1 ##. So each point on the line has coordinates of the form ## (t-1,3t,-t) ## and the tangent vector to this line is simply the derivative of the latter expression w.r.t. t which is ## (1,3,-1) ##.
My class has not learned anything about using derivatives with vectors is there another way to solve this without using derivatives?

thanks
 
  • #6
2,788
586
My class has not learned anything about using derivatives with vectors is there another way to solve this without using derivatives?

thanks
What textbook are you using?
I think its better for me to take a look at it to see what tools you have at hand.
 
  • #7
52
0
What textbook are you using?
I think its better for me to take a look at it to see what tools you have at hand.
This is the example we had from class
IMG_8274.jpg

IMG_8275.jpg
 
  • #8
2,788
586
Can you write the equation of those two lines in the form ## (x,y,z)=\vec a + \vec b t ##? This is the form in which the equation of the line is given in your example from the class.
 
  • #9
52
0
Can you write the equation of those two lines in the form ## (x,y,z)=\vec a + \vec b t ##? This is the form in which the equation of the line is given in your example from the class.
(x,y,z)=(-1/2,-1/2-5/2)+t(1/2,1/2,1)
and
(x,y,z)=(-1,0,0)+s(1,3,-1)

would I have to combine these?
 
  • #10
2,788
586
(x,y,z)=(-1/2,-1/2-5/2)+t(1/2,1/2,1)
and
(x,y,z)=(-1,0,0)+s(1,3,-1)

would I have to combine these?
Good. Now those constant vectors which are the coefficients of t and s are the tangent vectors to those lines. Now you can either take the vector product of those tangent vectors as the tangent vector for your line or consider a vector with unknown components and set its scalar product with both tangent vectors to zero and find the components and take it as the tangent vector for your line.
 
  • #11
52
0
Good. Now those constant vectors which are the coefficients of t and s are the tangent vectors to those lines. Now you can either take the vector product of those tangent vectors as the tangent vector for your line or consider a vector with unknown components and set its scalar product with both tangent vectors to zero and find the components and take it as the tangent vector for your line.
Thanks I got it now
 

Related Threads on Vector equation perpendicular to two equations

Replies
5
Views
1K
Replies
4
Views
5K
  • Last Post
Replies
1
Views
5K
Replies
6
Views
44K
Replies
2
Views
494
Replies
3
Views
648
Replies
11
Views
1K
Top