Find a linear function whose graph is the plane that intersects the xy plane along the line.
I have an exam in an hour and this one question is killing me! We enter our answers online so maybe I am just entering it wrong, but I can't for the life of me seem to get it right.. any help?
Find a linear function whose graph is the plane that intersects the xy plane along the line y=x+8 and contains the point (2,2,16)
Somehow, the solution shows
y= x + 8, we have b /= 0
-a/b = 1, -c/b=8
Since (2,2,16) lies on the place we can use the equation z=ax+by+c to get
16 = 2a + 2b + 16c
Solving the solution gives
a=2 b=-2 c=16
giving 2x - 2y + 16
I just don't get it!
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$\begingroup$Hint: pick any two points on the line. The cross product of the vectors from $(2,2,16)$ to the two points is a normal to the plane.
$\endgroup$ 1 $\begingroup$With the equation of the line of intersection in the $ \ xy-$ plane ( $ z = 0 $ ) written in "general form" as $ \ -x + y = 8 \ , $ we can construct an equation for the tilted plane, also in "general form", as
$$ -Kx \ + \ Ky \ + \ cz \ = \ 8K \ , $$
the constant $ \ K \ $ being introduced to allow us a little "freedom" in the construction for the moment. This plane must contain the point $ \ (2, 2, 16) \ , $ so we have
$$ -2K \ + \ 2K \ + \ 16c \ = \ 8K \ \ \Rightarrow \ \ 2c \ = \ K \ . $$
Since any non-zero value for $ \ K \ $ will do, we may simply choose $ \ c = 1 \ $ to write an equation for the plane as
$$ -2x \ + \ 2y \ + \ z \ = \ 16 \ \ . $$
(The problem statement appears to be asking for the form $ \ z = 16 + 2x - 2y \ . $ )
This approach is analogous to finding, for example, the constant term $ \ C \ $ in the general equation for a line in the plane $ \ Ax \ + \ By \ = \ C \ $ , for specified coefficients $ \ A \ \text{and} \ B \ , $ passing through a specified point $ \ ( x_0 , y_0 ) \ . $
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