find the equation to the plane through the point (-1,3,2) and perpendicular to the plane (x+2y+2z=5) and (3x+3y+2z=8)
To find the equation of a plane passing through a given point and perpendicular to another plane, we can utilize the following steps:
1. Find the normal vector of the given plane:
- Consider the coefficients of x, y, and z in the equation of the given plane. For the first equation, the coefficients are 1, 2, and 2, respectively. For the second equation, the coefficients are 3, 3, and 2, respectively.
- The normal vector of the plane is obtained by taking these coefficients as the components, resulting in a vector N1 = (1, 2, 2) for the first equation and N2 = (3, 3, 2) for the second equation.
2. Find the cross product of the two normal vectors:
- Take the cross product of N1 and N2 to obtain a vector that is perpendicular to both vectors. Let's denote it as N.
- N = N1 x N2
3. Use the obtained normal vector and the given point to form the equation of the plane:
- Now, we have the normal vector N and the point P(-1, 3, 2).
- Substitute the values of N = (a, b, c) and P = (-1, 3, 2) into the equation of a plane: ax + by + cz = d, where d is a constant to be determined.
- Substitute the values of P(-1, 3, 2) into the equation: a(-1) + b(3) + c(2) = d.
- Rearrange the equation in terms of a, b, c, and d to obtain the equation of the plane.
Let's follow these steps and find the equation of the plane through the point (-1,3,2) and perpendicular to the plane (x+2y+2z=5) and (3x+3y+2z=8).
1. Find the normal vectors of the given planes:
- For the first equation (x+2y+2z=5): N1 = (1, 2, 2)
- For the second equation (3x+3y+2z=8): N2 = (3, 3, 2)
2. Find the cross product of the two normal vectors:
- N = N1 x N2 = (1, 2, 2) x (3, 3, 2)
- N = (4, -2, -3)
3. Use the obtained normal vector and the given point to form the equation of the plane:
- Substitute the values P = (-1, 3, 2) into the equation: 4(-1) + (-2)(3) + (-3)(2) = d
- Simplify the equation: -4 - 6 - 6 = d
- Calculate d: d = -16
Therefore, the equation of the plane through the point (-1,3,2) and perpendicular to the plane (x+2y+2z=5) and (3x+3y+2z=8) is:
4x - 2y - 3z = -16