Question

What would the equation of the line normal to this surface be? x^2-y^2+z^2=4 at the point (2,-3,3)

Answer 1

I know the vector is <2x,-2y,2z> but how do I do the equation? @amistre64

Answer 2

how do we define a line?

Answer 3

y=mx+b?

Answer 4

no ... you are working with vectors here. think parametrically

Answer 5

2xi-2yj+2zk?

Answer 6

given a point (xo,yo,zo) and a direction vector (a,b,c)

the parametric form of a line is defined as:

x = xo + at
y = yo + bt
z = zo + ct

Answer 7

your gradient can be 'simplified' since the length of it is immaterial: factor our the 2 :)

Answer 8

oh I didn't know I could do that thank you

Answer 9

a direction is whats important to:

<2x,-2y, 2z> is in the same direction as: <x,-y,z> right?

Answer 10

right

Answer 11

so then x=2+xt, y=-3-yt, z=3+zt ?

Answer 12

well, almost.

you need to define the gradient by the point as well

Answer 13

<2x,-2y,2z> @ (2,-3,3) = <2(2),-2(-3),2(3)> = 2<2,3,3>

x = 2+2t
y = -3+3t
z = 3+3t

Answer 14

ohh i get it and so those are the equations of the normal line?

Answer 15

yes

Answer 16

ah got it and what if i wanted to do the equation for the tangent plane?

Answer 17

the gradient tells us the normal at a given point

the line therefore is just the parametricized setup

Answer 18

we have a normal, dot it to all the vectors (x,y,z) to (2,-3,3)

Answer 19

when a dot product is equal to 0, then the vectors are perpendicular

Answer 20

gotcha thank youuu!! :)

Answer 21

youre welcome

all vectors from (2,-3,3) to (x,y,z) = (x-2, y+3, z-3)

dot this with our gradient to get

(2,3,3).(x-2, y+3, z-3)=0

2(x-2)+3(y+3)+3(z-3)=0

simplify as wanted