Vector geometry help

  • Thread starter mathrocks
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  • #1
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I have a question about solving the following problem
"Find the line through (3,1,-2) that intersects and is perpendicular to the line x=-1+t, y=-2+t, z=-1+t."

My question is regarding the intersection part, I know how to find the equation of the line that is perpendicular to another line but not both perpendicular and intersecting.

thanks
 

Answers and Replies

  • #2
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I'd think your forgeting that the dot product of two perpendicular vectors is 0 think about it...
 
  • #3
609
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what?
dot product....
the answer is as easy as you can think

(3,1,-2)+t(1,1,1)

why the hell you need dot product?
 
  • #4
daster
mathrocks said:
My question is regarding the intersection part, I know how to find the equation of the line that is perpendicular to another line but not both perpendicular and intersecting.
If two lines are perpendicular to each other, don't you think they're already intersecting?
 
  • #5
695
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what?
dot product....
the answer is as easy as you can think

(3,1,-2)+t(1,1,1)

why the hell you need dot product?
And that line is perpendicular to x=-1+t, y=-2+t, z=-1+t?

If two lines are perpendicular to each other, don't you think they're already intersecting?
No, two perpendicular lines in R^3 don't necessarily intersect.
 
Last edited:
  • #6
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Muzza said:
And that line is perpendicular to x=-1+t, y=-2+t, z=-1+t?
Wouldn't (3,1,-2)+t(1,1,1) be correct for the line perpendicular to x=-1+t, y=-2+t, z=-1+t since the vector <1,1,1> is coming from that?
 
  • #7
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A direction vector of (3,1,-2)+t(1,1,1) is (1,1,1), and a direction vector of x=-1+t, y=-2+t, z=-1+t is (1,1,1). Two lines are perpendicular if and only if their direction vectors are perpendicular. Surely (1,1,1) is not perpendicular to (1,1,1)?
 
  • #8
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Muzza said:
A direction vector of (3,1,-2)+t(1,1,1) is (1,1,1), and a direction vector of x=-1+t, y=-2+t, z=-1+t is (1,1,1). Two lines are perpendicular if and only if their direction vectors are perpendicular. Surely (1,1,1) is not perpendicular to (1,1,1)?
So, how would you go about finding a line that is perpendicular?
 
  • #9
daster
Muzza said:
No, two perpendicular lines in R^3 don't necessarily intersect.
Ah.. I over-looked that.
 
  • #10
daster
mathrocks said:
So, how would you go about finding a line that is perpendicular?
The dot product between their direction vectors should be 0.
 
  • #11
HallsofIvy
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Note: there exist an infinite number of lines (in 3D) through a given point and perpendicular to a given line. Perhaps part of the problem is that there are too many answer. You should know that, in 2D, a perpendicular to <a, b> is <b, -a> since that way <a, b>.<b,-a>= ab-ab= 0. In 3d, the simplest thing to do is just take one of the components 0 so that the problem goes back to 2D.

To find a perpendicular to <1, 1, 1> try <0, a, b> What must a, b be?
Of course, <a, 0, b> and <a, b, 0> would work as well.
 
  • #12
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the line has the vector <1,1,1> as its direction
dot with the vector that lies on the perp line c...
[tex]
\vect{c} = <c_x, c_y, c_z>[/tex]
[tex]<1,1,1> dot <c_x, c_y,c_z> = 0 [/tex]
gives you
[tex]
c_x = -c_y - c_z [/tex]
[tex]c_y free [/tex]
[tex]c_z free
[/tex]

subracting the vector that points to the point on the line of perpendicularity (r), and the vector the point we are trying to make a line through gives us the vector in the direction of the line which is c...
[tex]
<r_x, r_y, r_z> - <3,1,-2> = <-c_y-c_z, c_y, c_z>
[/tex]
[tex]r_x - 3 = -c_y - c_z [/tex]
[tex]r_y - 1 = c_y [/tex]
[tex]r_z + 2 = c_z [/tex]

however the terminal point of vector r must lie on the origal line..there fore..

r_x = t -1
r_y = t - 2
r_z = t - 1 ...substute...

[tex]
(t-1) - 3 = -c_y - c_z [/tex]
[tex] (t-2)-1 = c_y [/tex]
[tex](t - 1) + 2 = c_z [/tex]

which we can easily solve for t...t = 2 put find the vector c by substuting in the t value...you'll get...
[tex] \vect{c} = <-2, -1, 3> [/tex]
now we can form the line equation
x = -2t +3
y = -t + 1
z = 3t -2

done...check the dot product its perp
 
Last edited:

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