/sci/ - Science & Math

[Anonymous No.2202879]

Wondering if anyone could solve this quickly for me

It was on my final today and I'm not sure if I got it right

anyway, decompose the vector 3i+2j into two vectors, v1 parallel to i-j, and v2 orthogonal to i-j

[Krakengineer !!5XY+x7grkpt No.2202922]

OK

v1 = i-j

v2 = i+j

x = 3i+2j

Shove all that into a matrix
{{x,v1,v2,i,j},{1,0,0,3,2},{0,1,0,1,-1},
{0,0,1,1,1}}
Do some matrix operations until I've got a row with zeros in everything except the x, v1, and v2 coloumns, and I get:

x = -(1/2)v1 - (5/2)v2

>> There's a chance I fucked up the matrix operations. I always hate doing those by hand.

[Anonymous No.2202944]

New base (i,-j) (-i,-j)

Since they are the same mathematical object

3i+2j=a(i,-j)+b(-i,-j)

3=a-b

2= -a -b

5=-2b b = -5/2

2a=1 a =1/2

New coordinates (1/2,-5/2)

[Anonymous No.2202979]

I'm not understanding either of these answers. You both had an end result of 1 vector, but you're supposed to have 2

What I was taught was to project the original vector onto the unit vector for v1, then to subtract v1 from v to get v2

[Anonymous No.2203049]

We're trying to find two vectors such that
(3,2) = a (1,-1) + (v1,v2)
where (v1,v2) . (1,-1) = 0 (dot product)
so v1 - v2 = 0, so v1 = v2

so (3,2) = (a,-a) + (v1,v1)
3 = a +v1
2 = -a +v1
solving the system equations gives
v1 = 2.5
a=0.5
so the vectors are: (2.5,2.5), which is orthogonal to i-j, and (0.5,-0.5), which is a multiple of i-j, so it is parallel to it.