Question

Find an equation of a plane.
The plane passes through the point(-1,2,1) and contains the line of intersection of the planes x + y - z = 2 and 2x - y +3z = 1
help please

Answer 1

Well, first you need to find the line where the two planes meet, can you do this?

Answer 2

i am not sure

Answer 3

So far I took the normals of the 2 planes and did the cross product where I got
<2, -5, -3> this is the direction vector.
I found a point by making z = 0 for both equations of the plane.
My newly found point is (1,1,0)
I plug this new point 2(x-1) -5(y-1) -3(z-0) =0
therefore 2x -5y -3z = -3 and this is not what they get

Answer 4

if you cross the normals, i believe you get the vector for the line

Answer 5

using the normal vector <a,b,c> and a point common to both planes (its on the line of intersection) we can form the line equation as:\[x=x_o+at\]\[y=y_o+bt\]\[z=z_o+ct\]

Answer 6

you tried z=0 to find a common point

x + y = 2
2x - y = 1
----------
3x = 3 ; x=1

when x=1, y=1

therefore the point (1,1,0) is common to both planes

Answer 7

lets see how your cross worked out

x 1 2 x: 3-1 = 2
y 1 -1 -y: 3+2 = -5
z -1 3 z:-1-2 = -3

that was good

Answer 8

sooo the equation for the line is
\[x=1+2t\]\[y=1-5t\]\[z=-1-3t\]

Answer 9

ugh, i used a vector as a point lol ... lets try that with the 1,1,0

Answer 10

\[x=1+2t\]\[y=1-5t\]\[x=0-3t\]
thats better

Answer 11

...and i see that im reading the final outcome wrong; its not asking for the line equation is it; but the plane that contains the vector and a given point.

Answer 12

Thats is right is has me going in circles

Answer 13

well, we only need 2 vectors to define a normal for the plane that contains them right?

Answer 14

one vector is <2,-5,-3>

the other is the vector from (1,1,0) to (-1,2,1)

Answer 15

so with our new found point where extract the other vector = <-2,1,1> then do cross product with <2,-5,-3>

Answer 16

(1,1,0)
-(-1,2,1)
--------
<2,-1,-1> ; or as you did <-2,1,1>

yes cross them and attach that normal to either point for the setup of the plane

Answer 17

The X product i get is <2,-4,8> then drop in any point say (-1,2,1) they get as final answer x -2y +4z = -1 which is different to where we are heading

Answer 18

<2,-4,8>; or simplified as: <1,-2, 4>

\[1(x-x_o)-2(y-y_o)+4(z-z_o)=0\]
\[(x-1)-2(y-1)+4(z-0)=0\]
\[x-1-2y+2+4z=0\]
\[plane:~x-2y+4z=-1\]

where are we heading?

Answer 19

That is brilliant, so you factored out 2. how do I give you a medal on this site?

Answer 20

to the right of my name there should be a "best response" option that you can click. if it is not there, you might have to refresh the screen.

Answer 21

Thanks very much for your help

Answer 22

youre welcome, and good luck ;)

Answer 23

What textbook are you using to get this stuff? I am using James Stewart Calculus