Parametric equation for a line

[ajoywatkins]

Hello,

So I am trying to solve this problem

Consider the line perpendicular to the surface z=x^2 +y^2 at the point where x = 4 and y = 2. Find a vector parametric equation for this line in terms of T

L(t) = ?

So first my goal was to find the tangent plane so I could the normal vector.

Tangent plane:

Partial derivative for x is 2x and partial derivative for y is 2y, so the gradient is

< 2x, 2y>

and the gradient at the point x = 4 and y = 2 is

<8,4>

So the equation for the plane is

z = 20 + 8(4-x) + 4(2-y)

Then I need two vectors in the plane

So points we know are on the plane are

P = (4,2,20)
Q = (1,1,48)
R = (1,2,44)

So I found the vectors

PQ = <-3,-1,28>
PR = <-3,0,24>

To find the normal vector, I calculated the cross product (Using a cross product calculator just in case this was my area of error)

I got <-24,-12,-3>

So my equation

L(t) = (4,2,20) + t<-24,-12,-3> (I don't need to reduce the vector as the system does that on its own)

The system kicks back and tells me this is wrong. I'm not sure where I am going wrong.

 

[galactus]

I think you just have some wrong signs.

If you rewrite to \(\displaystyle z-x^{2}-y^{2}=0\), then find the partials:

\(\displaystyle F_{x}=-2x, \;\ F_{y}=-2y, \;\ z=1\)

Plugging in x=4, y=2, z=20 the normal line has direction numbers:

-8i-4j+k

leading to:

\(\displaystyle (4,2,20)+t(-8,-4,1)\)

which is what you have except for the signs.

 

[ajoywatkins]

I think you just have some wrong signs.

If you rewrite to \(\displaystyle z-x^{2}-y^{2}=0\), then find the partials:

\(\displaystyle F_{x}=-2x, \;\ F_{y}=-2y, \;\ z=1\)

Plugging in x=4, y=2, z=20 the normal line has direction numbers:

-8i-4j+k

leading to:

\(\displaystyle (4,2,20)+t(-8,-4,1)\)

which is what you have except for the signs.

Thank you so much Galactus!

I feel rather silly now for doing all the extra work on top of it.