Vectors and Parametric Curves

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  • #1
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Homework Statement


Consider the line and plane below.
x = 5-5t, y = 3+7t, z = 10t
ax + by + cz = d

Find values of a, b, c, and d so that the plane is perpendicular to the line and through the point (2, 1, 2).

Homework Equations


Fgrad=(x',y',z') is perp to surface
if [tex]\vec{v}[/tex]1[tex]\bullet[/tex][tex]\vec{v}[/tex]2=0
then v1[tex]\bot[/tex]v2


The Attempt at a Solution



when I put those x,y,z values together I get a parametric equation that equals <5,3,0>+t<-5,7,10>

the starting point <5,3,0> I think is not relevant to finding a perpendicular vector,
I just need to find a vector perpendicular to t<-5,7,10> that goes through (2,1,2) i think? but i'm not sure how to do this..
then once I find the vector thats perp to t<-5,7,10>(dot)<a,b,c>=0
once I know a,b,c I can solve for d
 

Answers and Replies

  • #2
Consider the equation of the line:
(x,y,z) = (5,3,0) + t(-5,7,10)
Which is really just translated from:
(x,y,z) = t(-5,7,10)
Which is what you have so far.
So by definition the plane is defined by all vectors perpendicular to (-5,7,10)
Which is just 7y + 10z - 5x = 0
Translated plane:
v dot (x-x0) = 0
You complete.
 
  • #3
HallsofIvy
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Homework Helper
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Homework Statement


Consider the line and plane below.
x = 5-5t, y = 3+7t, z = 10t
ax + by + cz = d

Find values of a, b, c, and d so that the plane is perpendicular to the line and through the point (2, 1, 2).

Homework Equations


Fgrad=(x',y',z') is perp to surface
if [tex]\vec{v}[/tex]1[tex]\bullet[/tex][tex]\vec{v}[/tex]2=0
then v1[tex]\bot[/tex]v2


The Attempt at a Solution



when I put those x,y,z values together I get a parametric equation that equals <5,3,0>+t<-5,7,10>

the starting point <5,3,0> I think is not relevant to finding a perpendicular vector,
I just need to find a vector perpendicular to t<-5,7,10> that goes through (2,1,2) i think?
No, you don't. You want to find a vector perpendicular to the plane in order to write the equation of the plane. Since the line itself is perpendicular to the plane, its "direction vector", <-5, 7, 10>, is perpendicular to the plane.
So you know the plane can be written as [itex]-5(x-x_0)+ 7(y-y_0)+ 10(z-z)= 0[itex], where [itex](x_0, y_0, z_0)[/itex] is some point in the plane. Any you are also given that.

but i'm not sure how to do this..
then once I find the vector thats perp to t<-5,7,10>(dot)<a,b,c>=0
once I know a,b,c I can solve for d
 
  • #4
13
0
Ok, ok I understand now

since the line is already perp. to the plane, the direction of the line corresponds to the gradient of the plane which is <a,b,c> and then i just use the equation of a plane to solve
d=a(x-xo)+b(y-yo)+c(z-zo)

thanks
 

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