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March 28th, 2016, 10:02 AM   #1
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Determine a scalar equation

Determine a scalar equation for the plane that passes through the point (2,0,−1) and is perpendicular to the line of intersection of the planes
2x + y – z + 5 = 0 and x + y + 2z + 7 = 0.

I know that the line of intersection is in both planes, so its direction vector must be perpendicular to both normal vectors (2, 1, -1) and (1, 1, 2).

But I'm not sure where to go from there.

Last edited by skipjack; March 28th, 2016 at 01:26 PM.
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March 28th, 2016, 12:03 PM   #2
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And that direction vector must be perpendicular to the desired plane.

A plane with normal vector <A, B, C>, containing point $\displaystyle (x_0, y_0, z_0)$ has equation $\displaystyle A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$.
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March 28th, 2016, 01:21 PM   #3
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Quote:
Originally Posted by puppypower123 View Post
I know that the line of intersection is in both planes, so its direction vector must be perpendicular to both normal vectors (2, 1, -1) and (1, 1, 2).

But I'm not sure where to go from there.
The vector cross product $\vec a \times \vec b$ is a vector that is perpendicular to both $\vec a$ and $\vec b$.
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Last edited by skipjack; March 28th, 2016 at 01:27 PM.
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March 28th, 2016, 01:29 PM   #4
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So would i place it like this to get the solution?
A(2−1)+B(1−1)+C(-1−2)=0
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March 28th, 2016, 01:34 PM   #5
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Oh wait! I understand! I use both!

n⃗ =n⃗ 1×n⃗ 2=(2,1,−1)×(1,1,2)=(3,−5,1)
(x−2,y,z+1).(3,−5,1)=0 ⇒ 3x − 5y + z = 5

Last edited by skipjack; March 28th, 2016 at 02:22 PM.
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March 28th, 2016, 02:14 PM   #6
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That looks OK to me.
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March 31st, 2016, 09:50 AM   #7
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Quote:
Originally Posted by puppypower123 View Post
So would i place it like this to get the solution?
A(2−1)+B(1−1)+C(-1−2)=0
A(x- 1)+ B(x- 1)+ C(x- 2)= 0.
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