find two perpendicular vectors u and v such that each is perpendicular to w = <-4, 2, 5>
this smells like the grahm schmit process which i don't remember.
assuming you're using the standard dot product denoted by a period.
for the first vector, you want <a,b,c>.<-4, 2, 5>=0.
how about a=3, b=1, c=2? <3,1,2>.<-4,2,5>=-12+2+10=0.
for the second vector, take the cross product denoted by x of those two:
vector 3 = <3,1,2>x<-4,2,5>. i get <1,-23,10>.
how about not first assuming a=3, b=1 and c=2, but instead using equations to figure out the vectors
well, you can let <a,b,c> and <d,e,f> be two vectors meeting your criteria.
then <-4, 2, 5>.<a,b,c>=0, <-4, 2, 5>.<d,e,f>=0, and <a,b,c>.<d,e,f>=0.
that takes more work than an ordinary linear system i think.
No of Unknows a,b,c,d,e,f and no of eq 3.
?U cant find it unless some additional conditions are given
there are infinitely many solutions.
if you take the following values, note that it works:
Yes u can find a relation not a unique solution
"find two perpendicular vectors u and v such that each is perpendicular to w = <-4, 2, 5>"
Since the problem only asks for two vectors and there are infinitely many, I would do it this way:
Write u= <2, 4, 0>. By taking the k component 0, I can just "swap" the i, j components to get <2, 4, 0>.<-4, 2, 5>= 0 trivially.
Now take the cross product of those two vectors to find a third vector orthogonal to both. I get <20,-10,20> and since we only want "perpendicular", <2, -1, 2> will do.
u= <2, 4, 0> and v= <2, -1, 2> are perpendicular to each other and both are perpendicular to <-4, 2, 5>.
so the common thread seems to be to 'eyeball' a solution in the first case and then to use the cross product to find the third vector. if the question was as stated then that's what you're supposed to do. now if you haven't learned about the cross product yet, then here's what you might do:
0. start with w=<-4,2,5>.
1. eyeball a vector whose dot product with w is 0. for example, u=<2,4,0> works. (you are doing one more thing than just swapping the coordinates though; you're also switching one sign.)
2. now go to equations. let v=<a,b,c>. we want v.w=0 and v.u=0.
then -4a+2b+5c=0 and
to solve this system, note that you can solve the second equation to get a=-2b. then when you substitute that into the first equation, you get -4(-2b)+2b+5c=0 or 10b+5c=0. from this, you get b=-c/2. going back to a=-2b, you get a=c. note that you can't solve for all variables to get just one vector. that's because this set of equations has only two in it while there are three unknowns.
so v=<c,-c/2,c> satisfies the two conditions v.w=0 and v.u=0 for every value of c.
note that if you let c=20 or c=2, you get the two solutions hallsofivy produced.
this is more work than taking the cross product, of course.
1) Start with the given vector to which you want all others to be perpendicular. Normalize it (Not really necessary, but useful for step 2).
2) Take any other vector and subtract from it its own projection along the first vector (absolutely, completely, and by all means arbitrary, no eyeballing required (well, except, of course, that it can't be a scaled version of the first vector)). Normalize the result (Again, not necessary, but useful for step 3).
3) Take any other vector that is not a scaled version of the previous two, and subtract from it its own projection along the previous two vectors. ...
To make it more concrete, an example:
1) Let's say that we're starting with w. That takes care of step (1). But, for posterity, let's normalize:
w' = w/w = <-4,2,5>/5
2) Just as a randomly selected vector, I will take v'' = <1,0,0>. Nice and simple. Now I will subtract its projection:
v' = v'' - (v''.w')w'
= <1,0,0> - (-4/5)(<-4,2,5>/5) = <1,0,0> + <-16/25,8/25,4/5> = <9/25,8/25,4/5>
v = <9,8,20>/(√545)
(At this point, the advantage of initial eyeballing is aparent.)
3) To find the last vector:
u'' = <0,1,0>. Subtracting the projection: ...
Oh hell, I don't want to do it. Suffice it to say that you don't have to eyeball. Hopefully that makes you more comfortable with the procedure. Also note that you don't have to normalize either, but it makes the projection a simpler calculation.
I just thought of an alternative approach. Find the matrix that either rotates w to be along the z axis. Then, operate the inverse of the matrix on <1,0,0> and <0,1,0>. Does anyone know if this would work? I'll see if the matrix is too difficult.
OK, I can't remember. Given two matrices, Ω and Λ, and a vector, φ, does this relationship hold:
Ω(Λφ) = (ΩΛ)φ?
Is it possible to find a unique solution given the following cross products/conditions:
y x Z = X, Z x X = Y, and X x Y = Z?
This results in:
which I believe should be enough to solve this system (9 eq. and 6 unknowns). Is my reasoning correct?
Choose any vector, other than a multiple of w, for u and let v be the cross product of u and w.
We can do it as follows:
Let the original vector be (x,y,z). Write it in spherical polar coordinates as
(R sin theta cos phi, R sin theta sin phi, R cos theta).
Consider this vector to be obtained by starting with the (0,0,R), rotating by theta in the z-x plane and then by phi in the x-y plane. If we now take (R,0,0) and (0,R,0) and perform the same rotations, we'll get 2 new vectors,
(R cos theta cos phi, R cos theta sin phi, -R sin theta)
(-R sin theta, R cos theta, 0)
You can see that these are perpendicular to each other and to the original vector. In terms of the original components, we can write them
(zx/r, zy/r, -r)
(-Ry/r, Rx/r, 0)
where r = sqrt(x^2 + y^2)
the question is... |r(t)| = 1 for all t, then |r(t)| is orthogonal to r(t) for all t.
Take any vector that is perpendicular to w such as v=(5,0,4). Now have the third vector as u=(x,y,z) and solve the two equations u.v=0, u.w=0 (ofcourse you will not have a unique answer). From the last equation you get infinitely many solutions which infact is the vector space spanned by the third orthonormal vector. Just pick one such vector and you are done.
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