Question

Write the vector u=(2,3,-1) as the sum of two vectors, one parallel to v=(1,0,-3) and the other orthogonal to v=(1,0,-3)

Answer 1

I know I cna use projection, but Im not allowed in this case, I have to use sets of equations.

Answer 2

I can use proj_v_(U)

Answer 3

Ohk

Answer 4

Sorry! I just have to use something like this:

w is orthogonal, therefore w1*1+w2*0+w3*-3=0

and v=kz , therefore v=kz1,kz2,kz3. But I get stuck after this. I don't know how to make this into a system of equation where I can solve for a unique solution

Answer 5

The parallel vector is the magnitude of u=<2,3,-1> times a unit vector in the direction of v=<1,0,-3>

Answer 6

The perpendicular vector would then be u=<2,3,-1> - <the parallel vector>

Answer 7

suppose \(\mathrm{v\perp w}\) and \(\mathrm{u = av + bw}\)

Answer 8

from the dot product formula you have direction vector \(\mathrm{w = \langle 3,c,1\rangle }\)

yes ?

Answer 9

you get below equation
\[\large \langle 2,3,-1\rangle = a\langle 1,0,-3\rangle + b\langle 3,c,1 \rangle \]

Answer 10

3 equations and 3 unknowns
we can solve them

Answer 11

well, what do I do with the b parameter? put it as 1? If so, what is the point of inserting it in the equation? what about the c, how do we find that number?

I would have

2=a+3b
3=a0+bc
2=-3a+1b.

2+3a=b

so we can

2=a+2+3a
4=4a
1=a

So we can find b...
2=-3(1)+b
2+3=b
5=b

from there find c ( I assume
3=5c --> 3/5=c

so it would give vector 1 = (1,0,-3) and vector 2-> (15,3,5)
which does work. However I still do not understand why you put a c for the y variable of the second vector, which is what I was doing wrong all along

Answer 12

c is a free variable

Answer 13

w1*1+w2*0+w3*-3=0

solving this gives
w1 = 3w3
w2 = w2

yes ?

Answer 14

ah yeah, you can put it as anything, I see that. but if it's a free variable, then that means that it could be anything no ?

Answer 15

yep it could be anything as far as orthogonality is concerned

Answer 16

(1,0,-3) is perpendicular to (3k, ck, k)

no matter what the values of c and k are

Answer 17

rewrite it as (1,0,-3) is perpendicular to k(3, c, 1)

Answer 18

but then, when you plugged it in the system of equation above, c gets a single value and if its changed, its wrong, so it seems kinda contradicting to me!

Answer 19

c gets a single value because only one specific vector works for the projection

Answer 20

notice that the set of vectors perpendicular to (1,0,-3) form a plane

Answer 21

any vector in that plane is perpendicular to \(\mathrm{v}\), yes ?

Answer 22

It makes sense, so couldn't you use spanning to find the answer to this question then?

Answer 23

we represent all those vectors in that plane as : k(3, c, 1)

Answer 24

OH I see, I see it makes sense, but for that particular problem for the projection onto the other parallel vector, we need the c=5/3 . But other solutions would be poss8ible then, am I right? I mean the one I got here isn't set in stone?

Answer 25

there can only be one vector vector, let me show you why