Find an equation of the plane through the point 3 4 4 and pe

Find an equation of the plane through the point (3, -4, 4) and perpendicular to the vector (3, -4. -2). Do this problem in the standard way or WebWork may not recognize a correct answer. x + y + z =

Solution

The equation of a plane in R³ with  a point r₀ that lies in the plane and n, a vector normal to the plane,is n.( r –r₀ ) = 0 where r = (x, y, z)

Now, n (r – r₀) = 0 n r - n r₀ = 0 or, n r = n r₀ . On substituting the given values, we get   the following equation for the plane in Cartesian form: (3, -4, -2). (x, y, z) = (3, -4, -2). (3, -4, 4) or, 3x - 4y - 2z = 9 + 16 - 8 or, 3x - 4y – 2z = 17