Q. Our surface S is part of the plane containing the three points (1,0,0),(0,9,0), and (0,0,9). The equation of this plane isz=
Find Normal Vector: Step 1: Identify the normal vector to the plane using the given points.To find the normal vector, we use two vectors in the plane and find their cross product. The vectors can be formed using the given points: vector AB from (1,0,0) to (0,9,0) and vector AC from (1,0,0) to (0,0,9).Vector AB = (0−1,9−0,0−0) = (−1,9,0)Vector AC = (0−1,0−0,9−0) = (−1,0,9)Cross product of AB and AC = \begin{vmatrix}i & j & k\-1 & 9 & 0\-1 & 0 & 9\end{vmatrix}= i(9⋅9−0⋅0)−j(−1⋅9−0⋅(−1))+k(−1⋅0−9⋅(−1))= (0,9,0)0Normal vector = (0,9,0)1
Write Plane Equation: Step 2: Write the equation of the plane using the normal vector and a point on the plane.The general form of the plane equation is Ax+By+Cz=D, where (A,B,C) is the normal vector and (x,y,z) is any point on the plane.Using the normal vector (81,9,9) and point (1,0,0), substitute into the plane equation:81×1+9×0+9×0=D81=DSo, the equation of the plane is 81x+9y+9z=81
Simplify Plane Equation: Step 3: Simplify the equation of the plane.Divide the entire equation by 9 to simplify:981x+99y+99z=9819x+y+z=9Thus, the simplified equation of the plane is z=9−9x−y
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