Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
zonazoo
Group Title
For which scalars a, b, c, d do the solutions to the equation ax+by+cz=d form a subspace of R(3).
 one year ago
 one year ago
zonazoo Group Title
For which scalars a, b, c, d do the solutions to the equation ax+by+cz=d form a subspace of R(3).
 one year ago
 one year ago

This Question is Closed

vf321 Group TitleBest ResponseYou've already chosen the best response.0
In \(ax + by + cz = d\), if x and y are any number in R(1), then z is constrained to \[z=\frac{daxby}{c}\] The subspace S = { any (x, y, z)} is a plane which does not pass through (0,0,0) and is therefore not a subspace.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
Actually, I failed to address all the trivial cases. If a = b = c = 0, there is no solution. The empty set is not a subspace. If one or two of a,b, or c equals 0, then you can rotate your axes such that the one that is not 0 is the constant for z, in which case the theorem holds.
 one year ago

zonazoo Group TitleBest ResponseYou've already chosen the best response.0
so your saying that if atleast a, b, or c is 0 then the others can be non zero numbers and be a subspace of R(3)?
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
No, I'm saying that if d = 0 then the solution set (x, y, z) is a subspace of R(3) UNLESS all three are equal to 0.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.0
ll three a, b, c****
 one year ago

zonazoo Group TitleBest ResponseYou've already chosen the best response.0
and that would be the same thing if it was a, b, c to the equation ax+by=c form a subspace of R(2)... then c=0 in this case.
 one year ago

zonazoo Group TitleBest ResponseYou've already chosen the best response.0
also, if d=0, then as long as atleast one of a, b, or c is non zero then it IS a subspace right?
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
actually the set u r given is a subspace if and only if d=0.
 one year ago

zonazoo Group TitleBest ResponseYou've already chosen the best response.0
but why? and does it matter what a, b, and c are?
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
u have \[ \large D=\{(x,y,z):ax+by+cz=d\} \]
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
if d=0 then u have \[ \large D=\{(x,y,z):ax+by+cz=0\}. \] So let \(u=(x_0,y_0,z_0)\), \(v=(x_1,y_1,z_1)\) both in \(D\) and \(\alpha\in\mathbb{R}\) so (i) \(u+v=(x_0+x_1,y_0+y_1,z_0+z_1)\) so \[ \large a(x_0+x_1)+b(y_0+y_1)+c(z_0+z_1)= \] \[ \large =(ax_0+by_0+cz_0)+(ax_1+by_1+cz_1)=0+0=0 \] (ii) \(\alpha v=(\alpha x_1,\alpha y_1,\alpha z_1)\) so \[ \large a(\alpha x_1)+b(\alpha y_1)+c(\alpha z_1)=\alpha(ax_1+by_1+cz_1= \alpha0=0 \] this proves that D is a subspace of R^3.
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
For the converse. Let's suppose that \[ \large D=\{(x,y,z):ax+by+cz=d\} \] is a subspace of R^3. then the zero vector MUST be in D so \[ \large 0=a0+b0+c0=d \]
 one year ago

zonazoo Group TitleBest ResponseYou've already chosen the best response.0
I do understand that for sure. but how did you know to make d=0, or how do you know it doesn't work for any non zero number?
 one year ago

zonazoo Group TitleBest ResponseYou've already chosen the best response.0
ohhh okay.
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
therefore D is a subspace if and only if d=0
 one year ago

zonazoo Group TitleBest ResponseYou've already chosen the best response.0
so that same idea applies if the question is ax+by=c, for scalars a, b, c ... subspace of R(2) right
 one year ago

zonazoo Group TitleBest ResponseYou've already chosen the best response.0
It seems like the book is trying to trick me... why give one question stating for ax+by=c, and then the question right after that be ax+by+cz=d, when c has to be 0 for the first and d has to be 0 for the second. Its the same question isnt it, or is there more to it that I am missing.
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
yes it is. the book is trying to stress that for a subset of a given to be a subspace (i.e., a space on its own) the zero vector (which is unique) must also belong to the subset in question.
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
that's what u r supposed to get from this two exercises.
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
*given space*
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
i hope i was helpful gotta go
 one year ago

zonazoo Group TitleBest ResponseYou've already chosen the best response.0
it most definitely was... i hope to see you around again, you have been very helpful. thank you.
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.