Can I Use Coefficients to Find the Normal Vector of a Plane?

[tony873004]

From the book's example, the normal vectors of the planes x+y+z=1 and x-2y+3z=1 are <1,1,1> and <1, -2, 3>.

Although the book doesn't mention how it got those normal vectors from the equations, it's rather obvious. But the first homework problem has the plane equation = 0 instead of equal 1. Can I still just pull the coefficients of x, y, z and form a normal vector? i.e. If the equation of the plane is x+z=0, then is the normal vector <1,0,1>?

 

[EngageEngage]

Yup, you can always pull the coefficients off for the normal because that term after the = sign doesn't change the slopes of the plane -- it will just determine intercepts and points through which the plane passes

 

[tony873004]

thanks! The book failed to explain that.

 

[tony873004]

I better post the full problem because I'm stuck again.

Find the parametric and symmetric equations of the line of intersection of the planes x+y+z=1 and x+z=0.

I got the normal vectors, <1,1,1> and <1,0,1> and their cross product <1,0,-1> or i-k.

I set z to 0 and got x=0, y=1, z=0.

How do I form parametric equation out of this?? I know it's x=t, y=1, z=-t because this problem is nearly identical to one from lecture. But how did he do that step?

This would make the symmetric equations x/1=y-1/0=z/-1. But I can't divide by 0, can I?

 

[HallsofIvy]

I better post the full problem because I'm stuck again.

Find the parametric and symmetric equations of the line of intersection of the planes x+y+z=1 and x+z=0.

I got the normal vectors, <1,1,1> and <1,0,1> and their cross product <1,0,-1> or i-k.

I set z to 0 and got x=0, y=1, z=0.

How do I form parametric equation out of this?? I know it's x=t, y=1, z=-t because this problem is nearly identical to one from lecture. But how did he do that step?

This would make the symmetric equations x/1=y-1/0=z/-1. But I can't divide by 0, can I?

I'm not sure why you are worrying about vectors. I would just solve the two equations for two of the variables in terms of the third. Subtracting the third equation from the first, we get y= 1 From the third equation, z= -x. Taking x itself as parameter, we have x= t, y= 0, z= -t.

As for the "symmetric" equations, yes, the fact that y is constant causes a problem! The only "symmetric" are z= -x, y= 1.

 

[tony873004]

A belated thank you, Halls. I didn't notice your response until now. I guess I stopped monitoring this thread after I turned in the homework. I also got z=-x, y=1. Thanks for confirming that for me.