How the f-ck do u do this question (1 Viewer)

[blackbunny]

deternime all real numbers k so that the following pairs of vectors are othogonal

u=(k,1,-2) v=(k,-3,k)

 

[acmilan]

Um vectors isnt in 3 unit, especially when they are in 3 dimensions

 

[KFunk]

If they're orthogonal then you could probably deal with them in terms of a plane couldn't you? Then you could prove that m1= -1/m2 gradient sorta thing. That's the best I can offer without having done vectors.

P.S This really isn't 3-unit stuff, you should post this either on the 4-unit (extracurricular) board or on the Vic Specialist board.

 

[acmilan]

Vectors are orthogonal if their dot products = 0 ie v.u = 0 which makes them perpendicular. However it is beyond the 3 unit course so ill move it to 4 unit.

 

[acmilan]

Thats why its in the extracurricular sub-forum

 

[blackbunny]

tnx for u r usless help

 

[illucid]

Useless?.....

You got told how to solve it. It just means that where u.v = 0 then it is considered orthogonal (the lines are perpendicular)
...... Easy (beyond 3u but not too much so).

 

[maths > english]

Solution attached

 

[maths > english]

The attached thing is to help explain the first part.

btw 3D diagrams havent been in the 3u course for about 30 years

 

[KFunk]

This is just me being unecesarily picky but where you were writing 'respectfully' did you mean to use respectively as in: 'in the order mentioned '? I just haven't seen/heard respectfully used like that unless you were intending to make it a very polite proof .

 

[maths > english]

yep, i meant in the order mentioned

 

[bhavo]

there is an easier nmethod... much easier.

to prove the dot product = 0, all you hafta do is multiply all the co-effs of the position verctors.

in english, that translates to, make a postion vector of point u and v, and multiply the corresponding co-effs of the 2 vectors.

here it is:

u: ki + j -2k
v: ki -3k + kk

where i, j, k are the postion vectors along the 3 axis, x, y, z.

now as acmilan poiunted out, of vectors are perpendicular their dotproduct = 0.

so when u multiply the coeffs and add em up, u solve that eqn:

coeff of vector i: k x k
coeff of vector j: -3 x 1
coeff of vector k: -2 x k

adding up gives u k^2 -2k -3, and that = 0

therefore k= -1 or 3.

 

[maths > english]

i think my explanation may have appeared much more complicated than it needed to be because the formula:

cosθ=cosα₁cosα₂+cosβ₁cosβ₂+cosγ₁cosγ₂

is a general one normally applied without explanation

i explained the formula because the person who asked the question was probably a hsc student who had never seen it before and because i didnt want to encourage them to blindly apply formulas without an understanding of what they were doing

 

[m_isk]

(beyond 3u but not too much so).

not so much so?? this is waaaaaaaaaayyy beyond the 3u and probably slightly beyond 4 as well...

 

[bhavo]

yeah its uni stuff. at least i learnt for the first time in uni. good 'ole MATH1902 (linear algebra).

its quite simple once u use the formula, but the explanantion involves learning all the stuff before, esp about position vectors, and 3D space, and the all the angles relative to the 3 axis.

not to difficult actually, if they taught it in 4U, most of the students would have probably understood it easily.