Find an Equation of Plane Through P (3,3,1) Perpendicular to Planes

[Giuseppe]

Hello, can anyone help me with this problem?

Find an equation of the plane through P (3,3,1) that is perpendicular to the planes x +y = 2z and 2x +z =10

Thanks

 

[amcavoy]

First take the two planes you have. The normal to the first is the vector <1,1,-2>. The normal to the second is <2,0,1>. Since both of these will essentially be parallel to the plane you are looking for, you want to find the vector perpendicular to that plane. In other words, the binormal vector to <1,1,-2> and <2,0,1>. How do you do that in a 3 dimensional coordinate system? Try using the cross product, and assuming you know how to get the equation of a plane from the normal vector and a point, you shouldn't have any trouble arriving at an answer.

 

[tacman]

in advance!

Sure, I can help you with this problem! To find an equation of a plane, we need to know three things: a point on the plane and two vectors that are parallel to the plane. In this case, we are given the point P (3,3,1), and we can find the two vectors by looking at the normal vectors of the given planes. The normal vector of x + y = 2z is <1,1,-2>, and the normal vector of 2x + z = 10 is <2,0,1>. Since the plane we are looking for is perpendicular to both of these planes, the normal vector of our plane must be perpendicular to both of these vectors, which means it must be their cross product. So, we can find the normal vector of our plane by taking the cross product of <1,1,-2> and <2,0,1>:

<1,1,-2> x <2,0,1> = <1,-4,-2>

Now, we can use this normal vector and the given point P (3,3,1) to write the equation of our plane in the form Ax + By + Cz = D. Plugging in the values, we get:

1(x-3) -4(y-3) -2(z-1) = 0

Simplifying, we get:

x - 4y - 2z + 9 = 0

So, the equation of the plane through P (3,3,1) that is perpendicular to the planes x + y = 2z and 2x + z = 10 is x - 4y - 2z + 9 = 0. I hope this helps! Let me know if you have any further questions.