# Finding distance for a given value between vectors

1. Oct 30, 2015

### Lanniakea

1. The problem statement, all variables and given/known data
Let u = (2,k,1,-4) and v = (3,-1,6,-3)

For which values of k is the distance d(u,v) = 6?
2. Relevant equations

3. The attempt at a solution

6 = √((3-2)2 + (-1-k)2 + (6-1)2 + (-3-4)2)
6 = √(1 + 1 + k2 + 25 + 21)
6 = √(k2 + 48)
36 = k2 + 48
k = √-12

The answers seem to be 6k, -16k and -k or k= 2 and -4 (the answer sheet makes no sense and is mislabeled and unorganized horrifically) and I don't even have a remote idea how to get multiple solutions. I suppose my whole approach is wrong and I can't for the life of me understand what to do (I'm working without any textbooks)

Kind regards

2. Oct 30, 2015

### Student100

Remember that $d(u,v) = ||u-v|| = \sqrt{(u_1-v_1)^2+(u_2-v_2)^2 ... + (u_n - v_n)^2}$, so you're on the right track. Recheck your math.

3. Oct 30, 2015

### Lanniakea

I redid the working (both d(u,v) and d(v,u) which turn out to be the same so it seems correct this time) and get k = √8

I really have no idea what I'm doing wrong, can you specify where my mistake is? I have a lot more work to catch up on and I'm struggling with the basics...

and by ||u−v|| does this mean the two magnitudes?

So I could solve it by 6 = √(k2+21) - √(55) which gives me k = √2 which is also obviously wrong.

What exactly is my mistake? Am I messing up +/- signs somewhere?

4. Oct 30, 2015

### Student100

Your first expansion in the original problem description was wrong, you will get a quadratic which has two solutions.

||u-v|| means the length of some new vector $\vec{r}$ that is the resultant from the vector operation $\vec{u}-\vec{v}$

My hint is that the expansion of say $(k+1)^2$ is $(k^2+2k+1)$, not what you're doing. Or $(k+1)(k+1)=(k+1)^2$

5. Oct 30, 2015

### Lanniakea

Ah I haven't done any maths in so long I had even forgotten basic factorization, I also thought that (k+1)2 = k2 + 12 for some reason, what a sin of a mistake! Thank you very much my friend.