Find Equation of Plane perpendicular to line passing through a point


by jheld
Tags: equation, line, passing, perpendicular, plane, point
jheld
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#1
Jun14-09, 05:18 PM
P: 81
1. The problem statement, all variables and given/known data
Find an equation for the plane that is perpendicular to the line x = 3t -5, y = 7 - 2t, z = 8 - t, and that passes through the point (1, -1, 2).


2. Relevant equations
Equation of a plane: Ax + By + Cz = D
D = Axo + By0 + Cz0


3. The attempt at a solution
I am not sure how to get the line x, y and z into the vector form <A,B,C>
thinking...
1 = 3t - 5....t = 2
-1 = 7 - 2t...t = 4
2 = 8 - t...t = 6

But using that in the equation of a plane does not seem to work. A little confused :(
Answer is : 3x - 2y - z = 3
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rock.freak667
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#2
Jun14-09, 05:26 PM
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Visually think the line perpendicular to the plane, what does the direction of line and the normal of the plane have in common?


also if you can put the line in the form

[tex]\frac{x-a}{p}= \frac{y-b}{q} = \frac{z-c}{r} (=t)[/tex]
jheld
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#3
Jun14-09, 05:34 PM
P: 81
Okay, well doesn't that mean that the line and the normal vector are 90 degrees in difference?
I can visualize how they interact.
I made the symmetric equations...
x + 5/3 = y - 7/-2 = -z + 8 = t

rock.freak667
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#4
Jun14-09, 05:42 PM
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Find Equation of Plane perpendicular to line passing through a point


Quote Quote by jheld View Post
I made the symmetric equations...
x + 5/3 = y - 7/-2 = -z + 8 = t
right so, from this form, look at the form I posted above, <a,b,c> is a point on the line and <p,q,r> is the direction.




Quote Quote by jheld View Post
Okay, well doesn't that mean that the line and the normal vector are 90 degrees in difference?
I can visualize how they interact.
(line)
|
|_______________ (plane)
|
|
|
|
(line and plane at 90 degrees to each other)

How does the direction of the line relate to the normal of the plane?
jheld
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#5
Jun14-09, 06:03 PM
P: 81
Direction of the line is parallel to the normal of the plane; thus we have the cross-product being equal to zero, which makes sense since the equation of a plane = 0. From here I can use a vector cross-product or do the dot-product, correct?
rock.freak667
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#6
Jun14-09, 06:11 PM
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P: 6,210
Quote Quote by jheld View Post
Direction of the line is parallel to the normal of the plane
that is correct.

But you don't need the cross-product here, since the direction is parallel to the normal, you can use any scale factor*the direction as the normal as the plane. To make things simple, just use the scale factor as one.

So what would be the normal of the plane? Can you find the equation of a plane given the normal and a point on the plane?
jheld
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#7
Jun14-09, 06:19 PM
P: 81
The normal is defined by the parameter t in this case, right? So shouldn't the normal vector be <3, -2, -1> ?
From there...
n dot-product vector initial point to end point = 0
jheld
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#8
Jun14-09, 06:22 PM
P: 81
Or using A(x - x_0) + B(y - y_0) + C(z - z_0 = 0
where the _0 indicates the initial point and x, y, z the ending of the vector..
3(x - 1) -2(y + 1) - 1(z - 2) = 0
3x - 3 - 2y - 2 -z + 2 = 0
3x - 2y - z = 3

Thank for the help :)
rock.freak667
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#9
Jun14-09, 06:23 PM
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Quote Quote by jheld View Post
The normal is defined by the parameter t in this case, right? So shouldn't the normal vector be <3, -2, -1> ?
From there...
n dot-product vector initial point to end point = 0
You can just take the direction of line to be normal in this case. If you wanted you could have the normal as <6,-4,-2> or even <30,-20,-10>. You'd still get the same answer in the end.
HallsofIvy
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#10
Jun15-09, 05:04 AM
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Thanks
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In other words, what is the equation of a plane with normal vector <3, -2, -1> containing point (1, -1, 2)?


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