Question

Find a basis that includes u = {(1,5,4)} in R^3.

Answer 1

any base?

Answer 2

the easy way or the not-so-hard way?

Answer 3

that's the only given. both ways will do :)

Answer 4

do u know about inner products?

Answer 5

if so then just solve
\[ \large 0=(x,y,z)\cdot(1,5,4)=x+5y+4z \]

Answer 6

we haven't discussed about the inner products or maybe I just don't know anything about it. Its our quiz tomorrow on vector spaces, spanning, linear combination,linear dependence and independence, and basis. Got any tips? :)

Answer 7

choose a 2nd that is most convenient, say (1,0,0), then work on the third basis (a,b,c) using the first two (1,5,4) and (1,0,0).

Answer 8

yes. just pick any vector not parallel (a scalar multiple) to the one given. any vector

Answer 9

then u have two choices: (i) use the cross product u learned in multi-variable calculus to get the third vector, for instance if u choose (1,0,0) then
\[ \large (1,5,4)\times(1,0,0)= \]
is the third one

Answer 10

(ii) check the consistency condition for the system
\[ \large \left(\begin{array}{cc|c}
1 & 1 & a\\ 5 & 0 & b\\ 4 & 0 & c
\end{array}\right) \]

Answer 11

this last one is more suited with what your examination is going to be about.

Answer 12

then pick a vector that doesn't fullfil the condition