find the equation of the planes 5x-3y+4=0 and x+y-2z+10=0 and which are 1 unit distance from the origin.?
To find the equation of the planes that are 1 unit distance from the origin, we need to find the normal vectors of the planes.
For the first plane, the equation is 5x - 3y + 4 = 0. To get the normal vector of this plane, we can write the equation in the form Ax + By + Cz = D, where A, B, and C are the coefficients of x, y, and z, and D is the constant term. The coefficients A, B, and C represent the components of the normal vector.
So, for the first plane, the normal vector is (5, -3, 0).
For the second plane, the equation is x + y - 2z + 10 = 0. In the same way, we can write the equation in the form Ax + By + Cz = D to get the coefficients. So, the normal vector of the second plane is (1, 1, -2).
Now, we need to find the equation of the planes that are 1 unit distance from the origin. We know that the equation of a plane in the form Ax + By + Cz + D = 0 can also be written as dot product of the normal vector and the position vector of any point on the plane equals -D.
Let's call the equation of the first plane P1 and the equation of the second plane P2.
For P1, we have (5, -3, 0) dot (x, y, z) = -4.
Therefore, the equation of the first plane that is 1 unit distance from the origin is 5x - 3y = -4.
Similarly, for P2, we have (1, 1, -2) dot (x, y, z) = -10.
Therefore, the equation of the second plane that is 1 unit distance from the origin is x + y - 2z = -10.
So, the equations of the planes that are 1 unit distance from the origin are 5x - 3y = -4 and x + y - 2z = -10.