March 28th, 2016, 09:02 AM  #1 
Newbie Joined: Mar 2016 From: Canada Posts: 24 Thanks: 0  Determine a scalar equation
Determine a scalar equation for the plane that passes through the point (2,0,âˆ’1) and is perpendicular to the line of intersection of the planes 2x + y â€“ 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 12:26 PM. 
March 28th, 2016, 11:03 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
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 $\displaystyle A(x x_0)+ B(y y_0)+ C(z z_0)= 0$. 
March 28th, 2016, 12:21 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra  The vector cross product $\vec a \times \vec b$ is a vector that is perpendicular to both $\vec a$ and $\vec b$.
Last edited by skipjack; March 28th, 2016 at 12:27 PM. 
March 28th, 2016, 12:29 PM  #4 
Newbie Joined: Mar 2016 From: Canada Posts: 24 Thanks: 0 
So would i place it like this to get the solution? A(2âˆ’1)+B(1âˆ’1)+C(1âˆ’2)=0 
March 28th, 2016, 12:34 PM  #5 
Newbie Joined: Mar 2016 From: Canada Posts: 24 Thanks: 0 
Oh wait! I understand! I use both! nâƒ— =nâƒ— 1Ã—nâƒ— 2=(2,1,âˆ’1)Ã—(1,1,2)=(3,âˆ’5,1) (xâˆ’2,y,z+1).(3,âˆ’5,1)=0 â‡’ 3x âˆ’ 5y + z = 5 Last edited by skipjack; March 28th, 2016 at 01:22 PM. 
March 28th, 2016, 01:14 PM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra 
That looks OK to me.

March 31st, 2016, 08:50 AM  #7 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  

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