----------------------------------- jamonathin Mon Nov 14, 2005 11:51 am Algebra Help. ----------------------------------- Hey guys. I was away since last wednesday, and i just got back last night, and i've already fallen behind in algebra. We're on proofs now, and i dont realy understand what the question(s) is/are asking. My teacher isn't here today and i got this assignment from a subsitute that's due tomorrow. Here's what the first question asks. Prove whether the following set of vectors form a basis for three-dimensional space. u = (1, 3, 2), v = (-5, 2, 1), w = (3, -1, 2) And here's what the second question is. Write (-16, 13, 5) as a linear combination of the three vectors above. Use scalars 'a', 'b' and 'c' in your answer Now in this second question I'm guessing I need to go . . (-16, 13, 5) = au + bv + cw and solve for 'a', 'b', 'c'. But is that right? And I really dont remember the rules for a three dimensional space. Do they have to be all linearly dependant? or what. . I'm not neccesarily asking anyone to solve them for me :roll:, but any help is appreciated, or even a link to website that explains this crapola. Thanks in advanced :). ----------------------------------- Brightguy Mon Nov 14, 2005 3:18 pm Re: Algebra Help. ----------------------------------- Matrix row-reduction is very nice to use here, but I assume they haven't taught that yet? Anyway, here are some pointers: -A basis for a three-dimensional space (R3) is formed by any set of 3 linearly independent vectors in R3. -Vectors {u,v,w} are linearly independent if and only if au+bv+cw=0 has only the trivial solution (i.e., a=b=c=0, where a, b, c are scalars). -If you haven't used matrices yet, when solving these, just write out the 3 equations formed (from the vector equation) and solve for a, b, c. E.g., the first problem leads to the following system of equations: a-5b+3c=0 3a+2b-c=0 2a+b+2c=0 And now solve for a, b, c (by substitution, elimination, etc.) The second problem is almost the same except instead of 0, 0, 0 on the right hand side of the equations you'll have -16, 13, 5. ----------------------------------- GlobeTrotter Mon Nov 14, 2005 4:13 pm ----------------------------------- Another quick way to check whether three vectors form a basis for 3-D space is to take the box product of them. If it equals 0, they don't form a basis, and are thus linearly dependant. For example (1, 3, 2) . (-5, 2, 1) x (3, -1, 2) != 0 ----------------------------------- Paul Mon Nov 14, 2005 4:59 pm ----------------------------------- Aka the Triple Scalar Product. Generally questions like the second one, you go from 3 equations, 3 unknowns, to 2 equations, 2 unknowns, to 1 equation, 1 unknown, solve then substitute up the ladder lol. And I agree, I'm not comfortable either with Discrete, it just doesn't feel natural like calculus. Guess its more practice for me. Though it just feels like "hollow" knowledge, I don't "feel" it in my chest. ----------------------------------- Mazer Mon Nov 14, 2005 5:10 pm ----------------------------------- And I agree, I'm not comfortable either with Discrete, it just doesn't feel natural like calculus. Guess its more practice for me. Though it just feels like "hollow" knowledge, I don't "feel" it in my chest. Ah, can you elaborate on that? For me it's backwards. I never really loved the proofs, but I really liked the geometry stuff because it was easy for me to visualise and see a purpose for. ----------------------------------- jamonathin Mon Nov 14, 2005 7:52 pm ----------------------------------- Thanks for all the help guys, and I agree with Paul. Calculus comes together easier for me. Brightguy => Im not too shure about proofs still, so by substituting/eliminating those equations, that will solve the proof because i can find out if a=b=c=0 and in turn find out if they are linearly (in)dependant? GlobeTrotter => Good trick, ill use that to varify my answer. Ok now i see how the questions work, thanks again for your help guys. [It looked like it was gonna be harder. :)] ----------------------------------- jamonathin Mon Nov 14, 2005 7:58 pm ----------------------------------- Oh and just one more question (I just need a double check). This was the last question. If A divides BC internally in the ratio 2:5 and O is a point not on BC then express OA in terms of OB and OC So then this would be . . OA = 5/7OB + 2/7OC . . . and thats it? :? ----------------------------------- Brightguy Wed Nov 16, 2005 1:29 am Re: Algebra Help. ----------------------------------- Yep, but remember for the vectors to be linearly independent, it must have only the trivial solution. (Clearly, a=b=c=0 is always a solution to au+bv+cw=0). For #3, that looks good, although I guess technically there are 2 possible ways to split up a line segment into a 2:5 ratio...