Calculus- Early Transcendentals, 2021a

438 Three Dimensions
Working backwards, note that if (x,y,z) is a point satisfying ax + by + cz = d then
ax + by + cz = d
ax + by + cz − d = 0
a(x − d/a)+b(y − 0)+c(z − 0)=0
〈a,b,c〉·〈x − d/a,y,z〉 = 0.
Namely, 〈a,b,c〉 is perpendicular to the vector with tail at (d/a,0,0) and head at (x,y,z). This means
that the points (x,y,z) that satisfy the equation ax + by + cz = d form a plane perpendicular to 〈a,b,c〉.
(This doesn’t work if a = 0, but in that case we can use b or c in the role of a. That is, either a(x − 0)+
b(y − d/b)+c(z − 0)=0ora(x − 0)+b(y − 0)+c(z − d/c)=0.)
Figure 12.11: A plane defined via vectors perpendicular to a normal.
Thus, given a vector 〈a,b,c〉 we know that all planes perpendicular to this vector have the form ax +
by + cz = d, and any surface of this form is a plane perpendicular to 〈a,b,c〉.
Standard Form of a Plane
Anyplanecanbewrittenintheform
ax + by + cz = d
where a,b,c,d are constants and not all a,b,c are zero.
This plane is perpendicular to the vector 〈a,b,c〉.
Example 12.8: Perpendicular Plane
Find an equation for the plane perpendicular to 〈1,2,3〉 and containing the point (5,0,7).