Finding the equation of a plane

Asked Apr 01 '13 at 12:33

Active Apr 09 '13 at 03:05

Viewed 71 times

I need to find the equation for the plane which goes through the intersection line of the planes $x-z=1$ and $y+2z=3$, and perpendicular to the plane $x+y-2z=1$.

What I got so far - to find the intersection line I looked at $x-z = 1$ and $y+2x=5$, so I chose $z=0$ and I got a point $P_{1}$(1, 3, 0), which is on the intersection line.

Now I tried to express the plane equation using a point and a normal - I used $P_1$ as the point and $n_3 =<1, 1, -2>$ as the normal, so I got -

$x +y -2z = 4$, as the required plane.

Am I correct?

Thanks.

edited Apr 09 '13 at 03:05

J. M. ain't a mathematician

75,051

asked Apr 01 '13 at 12:33

homiee

Your equation defines a plane parallel to $x+y-2z=1$. Not perpendicular. – Julien Apr 01 '13 at 12:36
Both your plane and he one it is supposed to be perpendicular to have the same direction vector $,(1,1,-2),$...what do you think of that? – DonAntonio Apr 01 '13 at 12:36
What if I used $n_3 = <1,1,-1>$ instead? – homiee Apr 01 '13 at 12:51
1

It should be $n_3=<1,1,1>$. Here is a related problem. – Mhenni Benghorbal Apr 01 '13 at 12:58
How did you come up with $,n_3,$ , @homiee ? It is correct, BTW.(hint: inner product...) – DonAntonio Apr 01 '13 at 13:10
I searched for a vector $n_3$ which gives $n_3\cdot<1,1,-2>=0$ – homiee Apr 01 '13 at 13:11
@homiee: Notice that, I already gave a general technique for how to find such a vector formally in the link I referred you to it. – Mhenni Benghorbal Apr 01 '13 at 13:16
Thanks alot for your comments everyone – homiee Apr 01 '13 at 13:17

Finding the equation of a plane

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