anonymous
  • anonymous
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 y-axis and through the point (2, 3, 4) I know the x-axis is perpendicular to the y-axis, 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(x-x_0) + b(y-y_0) + c(z-z_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?
MIT 18.02 Multivariable Calculus, Fall 2007
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Ok, the vector equation of a plane is:\[n*(r-r_{0})=0\]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 y-axis. Therefore, our normal vector must be parallel to the y-axis. Any vector parallel to the y-axis will do, so I choose the unit vector <0,1,0> for simplicity. We have:\[<0,1,0>*=0\]giving,\[y=3\]This makes sense, since the y-coordinate must be constant, and there is no restriction on the values x,z that lie on the plane in 3-space.
anonymous
  • anonymous
This is great, thank you.
anonymous
  • anonymous
no worries man :)

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