Question

Determine the equation of the plane in the form
z = f(x,y) that is parallel to the vectors <9,6,2> and <−8,−4,−5,>,
and passes through the point (5,−5,8)

I am unsure where to start. I know I find the cross product, but after that I am lost.

Answer 1

So the cross product will be your normal to the plane. If you take that normal and dot it with the vector <x-x1,y-y1,z-x1> you should get the equation for the plane.

Answer 2

Where <x1,y1,z1> is a point in the plane.

Answer 3

you are given the point; P(x,y,z)

and the structure is to find the normal <a,b,c> and fill in the equation:

a(x-Px) +b(y-Py) +c(z-Pz) = 0

Answer 4

Err that should be <x-x1,y-y1,z-z1>. Oh, and the dot product should equal 0.

Answer 5

<9 , 6 , 2 >
x <-8,-4,-5>
-------------
< a , b , c >

Answer 6

a(x-Px) +b(y-Py) +c(z-Pz) <-- is the dot product I mentioned ;p

Answer 7

:) yes it is

Answer 8

ok, now i understand that the cross product will be normal to the plane but once i find the components of the cross product, do I subtract the points that it needs to pass through?

Answer 9

any given vector from a given point is determined by:

( x ,y , z )
-(Px,Py,Pz)
------------
which gives all vecotrs in a plane

Answer 10

Since any given vector in the plane can be found by a given point P(x,y,z)

then by doting all those vectors to the normal gives us the plane itself

Answer 11

given 2 points for example: (2,3,1) and (1,6,5); the vector between them is:

(2,3,1)
-(1,6,5)
-------
<1,-3,-4>; or < x0-x1 ,y0-y1 ,z0-z1 >

Answer 12

\[<a,b,c>\cdot<x-x_1, y-y_1, z-z_1> = 0\]\[\implies a(x-x_1)+b(y-y_1)+c(z-z_1)=0\]\[\implies ax + by + cz = ax_1 + by_1 + cz_1\]\[\implies <a,b,c>\cdot<x,y,z> = <a,b,c>\cdot<x_1,y_1,z_1>\]

Which that last one is usually how I solve these.

Answer 13

Less algebra that way.
\[<a,b,c>\] is the normal
\[<x_1,y_1,z_1>\]
is a point in the plane

Answer 14

all vectors from a point say (2,1,6) on the plane to any other point (x,y,z) are then:

( x ,y ,z )
-( 2 ,1 ,6 )
---------------
<x-2, y-1, z-6>

Answer 15

^ i see now thanks for your help

Answer 16

would the equation in terms of z be z = (22x - 29y +159)/12???

Answer 17

Generally the equation for a plane looks like:

ax + by + cz = d

Answer 18

i know, but i need to express it in terms of z for my hw.

Answer 19

that's odd, and pretty arbitrary but yes, it'd be (-dax -dby)/c then