Help with linear algebra

eoins23456

show that if a,d,c,d are real numbers and a,c,c are not all zero then az+by+cz=d is the equation of a plane in R^3 having the vector n=(a,b,c) as normal.

Not sure where to start.Im pretty sure we have covered it in lectures but i cant remember how to do it.Any idea where i should start?

sponsoredwalk

Was that post supposed to read:

eoins23456 wrote: »

show that if a,b,c,d are real numbers and a,b,c are not all zero then ax+by+cz=d is the equation of a plane in R^3 having the vector n=(a,b,c) as normal.

Not sure where to start.Im pretty sure we have covered it in lectures but i cant remember how to do it.Any idea where i should start?

???

Well the way I'd think about this is just to use the definition of a plane:

(X - P) • N = 0

where:
X = (x,y,z)
P = (p,q,r)
N = (a,b,c)

Read more about it here, part 6 of the first chapter. The picture isn't
showing up for me & the following pages are missing but basically P is a
point on the plane, X is another point anywhere you want on the plane &
is the variable with which you can create the whole plane centered at
point P (the vector arrow originates from P). The vector N is in the
direction perpendicular from the plane, it makes sense that a vector along
the plane (X - P) would be perpendicular to N so using the dot product set
equal to zero is just a smart way to work with it. Work with what I've
given you before looking in the spoiler, then you'll notice something very
general about equations of the form ax + by + cz = d, also try to work
backwards! All you need to remember is (X - P)•N = 0 and the rest is
easily rederived.

(X - P) • N = 0
X•N - P•N = 0
X•N = P•N
(x,y,z)•(a,b,c) = (p,q,r)•(a,b,c)
ax + by + cz = ap + bq + cr

ax + by + cz = d where d = ap + bq + cr

If a,b or c is zero then you'll end up with a term disappearing in the dot
product so they must all be nonzero.

eoins23456

Yeah typed that in a hurry.Thats clears up the question thank you.Did the dot product out and got what u got under the spoiler.thanks!