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

u = k * (1,0,-3) + t*orthogonal (1,0,-3)

Answer 2

Well, I just started by saying that the parallel vector ( let's say w) gives w1*1+w2*0-3*w3=0

Answer 3

but the orthogonal one... i dont know what to do with that one

Answer 4

orthogonal
(a,b,c) . ( 1,0,-3) = 0

Answer 5

oh crap, nvm parallel would be v2=kw

Answer 6

but where do I go from there, i have two system of equation with 3 unknown

Answer 7

so did we find an orthogonal vector

Answer 8

(a,b,c) . ( 1,0,-3) = 0

a*1 + b*0 + c *-3 = 0

pick a = 3 , b = 0, c = 1

(3,0,1) . (1,0,-3) = 0

Answer 9

orthogonal vector would be w1*1+w2*0-3*w3

Answer 10

why pick those random numbers? Seems kinda irelevant to me going with it as a guessing game

Answer 11

you can pick any values , it just has to be orthogonal

Answer 12

u = k * (1,0,-3) + m * ( 3 , 0 , 1 )

Answer 13

so we want to write u as a sum of a vector parallel to (1,0,-3) and a vector that is orthogonal to (1,0,-3)

Answer 14

if you didn't use the value m ( don't know why its there, seeing as it seemed that you establish the orthogonal vector as 3,0,1) you could use system of equation to solve for k right?

Answer 15

so knowing that u is 2,3,-1, you could say that for the x value, 3+k=2 therefore k = -1 right?

Answer 16

and do that for the 2 other values, if i understand correctly

Answer 17

i think so
solve
(2,3,-1) = k * (1,0,-3) + m * ( 3 , 0 , 1 )

Answer 18

but then the y value doesn't work because 0*k=3-0 therefore 0*k=3?

Answer 19

ok lets go back a step,

Answer 20

we can pick a simpler 'orthogonal' vector

Answer 21

we need a vector (a,b,c) that is orthogonal to (1,0,-3)
this means

(a,b,c) . ( 1,0,-3) = 0

a*1 + b*0 + c *-3 = 0

pick a = 3 , b =1, c = 1

(3,1,1) . (1,0,-3) = 3*1 + 1*0 + 1*-3 = 0

Answer 22

now we can solve

(2,3,-1) = k * (1,0,-3) + m * ( 3 , 1 , 1 )

Answer 23

2 = k + 3m
3 = 0k + 1m
-1 = -3k + 1m

so
3 = m

Answer 24

Wait, sorry if this sound stupid, but why would you put a coefficient in front of the orthogonal vector we found, I would think that we should find k and not m?

Answer 25

because any vector m*(3,1,1) is orthogonal to (1,0,-3)

Answer 26

there is not just one orthogonal vector, there are an infinite number of them , you can scale the orthogonal vector (and even flip the direction )

Answer 27

so I got
(2,3,-1) = -7 * (1,0,-3) + 3 * ( 3 , 1 , 1 )

check to see if that works

Answer 28

Ah , I see what you mean, the orthogonal vector can be multiplied by "infinite amount" of length, so I can use the coefficient m

Answer 29

right

Answer 30

The z value seems to be off -7*-3 + 3*1 = 24

Answer 31

(2,3,-1) = -7 * (1,0,-3) + 3 * ( 3 , 1 , 1 )

2 = -7*1 + 3*3
3 = -7 * 0 + 3*1
-1 = -7 * -3 + 3*1 , hmmm

Answer 32

lets go back to this step

(2,3,-1) = k * (1,0,-3) + m * ( 3 , 1 , 1 )

2 = k + 3m
3 = 0k + 1m
-1 = -3k + 1m

Answer 33

I can see that 1m=3 so I would assume m=3 but then Im stuck with the idea that k=-7.

Answer 34

ok this system does not work, we get a contradiction

Answer 35

this is trickier than i thought it would be :)

Answer 36

I know I'm sorry, I'm redoing my exam and I got that answer wrong, I still don't know how to approach it. it seems really tricky

Answer 37

im going to work on it , stand by