# Homework Help: Calc 3 Vectors

1. Sep 5, 2010

### kevin0788

1. The problem statement, all variables and given/known data
Let u= i-3j+2k, v=i+j, and w=2i+2j-4k. Find.
(d)||3u-5v+w||
(e)(1/||w||)*w
(f)||(1/||w||)*w||

2. Relevant equations
||V||=sqrt(v1^2+v2^2+v3^2)

3. The attempt at a solution
ok so for d i did 3(sqrt(1^2-3^2+2^2))-5sqrt(1^2+1^2)+sqrt(2^2+2^2-4^2)
and got 3sqrt(14) -5sqrt(2)+sqrt(8) but i know this is wrong because in the back of the book it says the answer is 2sqrt(37)
and for part e i tried (1/sqrt(2^2+2^2-4^2))*sqrt(2^2+2^2-4^2) and came up with 1/sqrt(32)*sqrt(32) and the back of the book says the answer is 1/sqrt(6)i + 1/sqrt(6)j - 2/sqrt(6)k.
i havent attempted part f yet. i thought it would be better to get part e straightened out first since they are similar. but could somebody point me in the right direction as to what i am doing wrong and maybe explain to me what the difference between part e and part f are. any help would be appreciated. thank you.

2. Sep 5, 2010

### Samuelb88

kevin0788 -

For part d, try computing the vector 3u-5v+w first, then determine the magnitude of the result. The difference between parts e, and d is in part e, you're finding the unit vector of w, and in part d, you're finding the magnitude of the unit vector of w.

- Sam

3. Sep 6, 2010

### HallsofIvy

||au+ bv|| is NOT equal to a||u||+ b||v|! As Samuelb88 said, first find 3u- 5v+ w, then find the norm.

No, that would be (1/||w||)||w||. First find 1/||w||= 1/sqrt{2^2+ 2^2+ (-4)^2}= 1/sqrt(24), not 1/sqrt(32), then multiply that number into each component of w.
The result is a vector, not a number.
Yes, after you have found the vector in (e), take its norm. However, it is true that, for a any non-negative number, ||av||= a||v|| so that ||(1/||w||)w||= (1/||w||)||w|| which is exactly what you did for (e)!