Linear Algebra(parametric equations)

[ur5pointos2sl]

The question states:
Determine the parametric equations of the line passing through the point P(2,1,0) and perpendicular to the plane 2x-5y=6.

The equation to obviously use is the "point-parallel" I'm guessing. X= P +tV where P is the point and V is a vector, and X=(x,y,z). But how would I use this since there isn't a vector given for V and instead is the plane 2x-5y=6? Maybe this isn't the one I need to use. Please help.

 

[Dick]

The direction vector for the line is the normal to the plane. Do you know how to find the normal to a plane?

 

[ur5pointos2sl]

The direction vector for the line is the normal to the plane. Do you know how to find the normal to a plane?

n . (x - p) ?

 

[Dick]

You can also write that as n.(x,y,z)=constant. What's a normal for your plane?

 

[ur5pointos2sl]

You can also write that as n.(x,y,z)=constant. What's a normal for your plane?

hm I honestly have no idea how to do this problem. I could set it up like this but then what would I do next?

N . (( x,y,z) - (2,1,0)= 0

x-2
y-1
z-0 ?

 

[Dick]

You don't need the point P to find the normal to the plane. Please look up how to find the normal to a plane ax+by+cz=constant.

 

[ur5pointos2sl]

You don't need the point P to find the normal to the plane. Please look up how to find the normal to a plane ax+by+cz=constant.

Ok I am not really finding too much on the topic.. I did however find one thing that said you want to use the cross product to find the norm to the plane.

 

[ur5pointos2sl]

would the norm happen to be (0,0,-4)

 

[Dick]

No, one choice for the normal would be (2,-5,0). Now where did I get that?

 

[ur5pointos2sl]

No, one choice for the normal would be (2,-5,0). Now where did I get that?

ahhh from the equation itself. so that would be the case everytime?

 

[JG89]

ahhh from the equation itself. so that would be the case everytime?

Whenever looking at the equation of a plane in the form ax + by + cz + d = 0, the vector (a,b,c) is always the vector that is perpendicular to the plane (which is a normal vector). Since you are trying to find the direction vector for a line perpendicular to the given plane, finding the direction vector is easy. You also have the point P, so you should be able to answer it now.

 

[Dick]

ahhh from the equation itself. so that would be the case everytime?

I'd hoped you could look this up but, yes, a normal to ax+by+cz=constant is (a,b,c). For your homework, explain to me why.