A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
For which scalars a, b, c, d do the solutions to the equation ax+by+cz=d form a subspace of R(3).
anonymous
 4 years ago
For which scalars a, b, c, d do the solutions to the equation ax+by+cz=d form a subspace of R(3).

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In \(ax + by + cz = d\), 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Actually, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 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)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and 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.

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

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1actually the set u r given is a subspace if and only if d=0.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but why? and does it matter what a, b, and c are?

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

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1if 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.

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1For 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 \[ \large 0=a0+b0+c0=d \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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?

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1therefore D is a subspace if and only if d=0

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It 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.

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1yes 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.

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1that's what u r supposed to get from this two exercises.

helder_edwin
 4 years ago
Best ResponseYou've already chosen the best response.1i hope i was helpful gotta go

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it most definitely was... i hope to see you around again, you have been very helpful. thank you.
Ask your own question
Sign UpFind more explanations on OpenStudy
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.