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anonymous
 5 years ago
I think this question is pretty straightforward and I'm making it more complicated than it needs to be.
Find the equation of the vertical plane perpendicular to the yaxis and through the point (2, 3, 4)
I know the xaxis is perpendicular to the yaxis, and in 3D space I know this must be the XZ plane. From here, I get a little lost coming up with the equation.
If the equation of a plane is a(xx_0) + b(yy_0) + c(zz_0) = 0, then my a = 2, b = 3, c = 4. But I feel like that's not right. Do I plug in those values for x, y, z instead?
anonymous
 5 years ago
I think this question is pretty straightforward and I'm making it more complicated than it needs to be. Find the equation of the vertical plane perpendicular to the yaxis and through the point (2, 3, 4) I know the xaxis is perpendicular to the yaxis, and in 3D space I know this must be the XZ plane. From here, I get a little lost coming up with the equation. If the equation of a plane is a(xx_0) + b(yy_0) + c(zz_0) = 0, then my a = 2, b = 3, c = 4. But I feel like that's not right. Do I plug in those values for x, y, z instead?

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, the vector equation of a plane is:\[n*(rr_{0})=0\]where n is a vector normal to the plane, r is the position vector of an arbitrary point in the plane and r0 is the position vector of a given point in the plane. We need a normal vector to solve this. To get a normal vector, we use the given information that our plane is perpendicular to the yaxis. Therefore, our normal vector must be parallel to the yaxis. Any vector parallel to the yaxis will do, so I choose the unit vector <0,1,0> for simplicity. We have:\[<0,1,0>*<x2,y3,z4>=0\]giving,\[y=3\]This makes sense, since the ycoordinate must be constant, and there is no restriction on the values x,z that lie on the plane in 3space.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is great, thank you.
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