Question

A plane that contains (3,-1,2) point and the line r(t)=<2,-1,0>. What is the equation of that plane?

Answer 1

Very useful notes: http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx (with great pics)

Nice video also: http://patrickjmt.com/finding-the-scalar-equation-of-a-plane/

Hope this helps. :)

Answer 2

thanks! I'll check them out right now.

Answer 3

oops* the line r(t) should be = <2,-1,0>+t<2,3,0>

Answer 4

I watched the video and went over the pauls notes section (both of which I've used and are great resources) but I am drawing a blank on the line part r(t).

Answer 5

I believe I need to do something to it to get a normal vector and then I cross that with the point given and then sub the values into the equation a(x-x0)+b(y-y0)+c(z-z0)=0

Answer 6

You just need to take two vectors on the plane and cross product that to get the normal.

Answer 7

<3,-1,2> X <2,-1,0> ?

Answer 8

Not quite, you can also think of in terms of points. Let A(3,-1,2). Take another two points on the line (by plug in two values of t, up to you which t) to get B and C. Now you have Vector BA and BC and you can cross product that.

Answer 9

is this on the right track? i got <2+2t,-1+3t,0> would this be correct to sub a number in for t?

Answer 10

correct you substitute two values of t to get two points on the line. Then from the three points you have two vectors.

Answer 11

so I got these three points t(0) B= <2,-1,0> , t(1) C=<4,2,0>

now I take that <3,-1,2> X <2,-1,0> => Vect BA
<2,-1,0>X<4,2,0> => BC
BA X BC => a point
take that point and sub the x,y,z values into the equation a(x-x0)+b(y-y0)+c(z-z0)=0

Answer 12

Remember to get the vector say BC you do <4-2,2-(-1),0-0>=<2,3,0>

Answer 13

hmm, For BC i got 0,0,8 when I <2,-1,0>X<4,2,0> I believe you used -1 for the sub while i used 0

Answer 14

I went through and got <32,-16,0> for the abc values and ended with:
32(x-3)-16(y+1)+0

Answer 15

Hod on ....
I used the points that you obtained earlier.
Remember that if A(a,b,c) and B(d,e,f)
Then the vector AB is given by <d-a,e-b,f-c>

Answer 16

ohhh, I was doing a+d. Let me retry

Answer 17

<2t-2,3t+1,2t-2> does this look better?

Answer 18

Ok, lets use (a,b,c) for point and <a,b,c> for vector to avoid some confusion.
Using the points A(3,-1,2) and B(2,-1,0) , C(4,2,0) that you obtained by plugin t=0,t=1 we have two vectors
BA=<3-2,-1-(-1),2-0>=<1,0,2>
BC=<4-2,2-(-1),0-0>=<2,3,0>
Then you cross product BA and BC to get the normal vector.

Answer 19

<-6,4,3> ?

Answer 20

BA X BC

Answer 21

yes