# Cakculating vectors

1. Sep 14, 2004

### ppacc

Question :
Knowing that Vectors C = (4x-1y-3z) meters and D=(2x-3y-5x) meters
Determine
a. S=C-D
b. │S
c. The Unitary vector in the direction of S

My Prof has given us a test one week after the first class and I just don’t get these study questions. I can add and multiply vectots but this stuff I just don’t get and there is noone to help.

Pamela

Last edited: Sep 14, 2004
2. Sep 14, 2004

### marlon

I don't get your notation. What are the base-vectors describing the directions ???

Something like 6e_x + 7e_y. The e_x and e_y denote the x and y direction and these are the base-vectors. 6 and 7 are the components so this vector may also be written as (6,7).

When adding or substracting just add or substract all components per direction. So for example (7,8) - (1,1) yields (6,7) or 6e_x + 7e_y

Multiplying means multiply the components per direction and add up the outcomes: so this means (7,8)*(1,2) = 7*1 + 8*2 = 7+16=23

If you take the squareroot of this you get the length of a vector = sqrt(23) like your question b.

But first of all you need to know the components per direction. What are these j and m???
Are you sure you got the notation right??? Just wondering

regards
marlon

3. Sep 14, 2004

### ppacc

Changes

Thanks Marlon I made the changes by editing my post. Maybe now my question is easier to understant

Pamela

4. Sep 14, 2004

### marlon

So you have (4,-1,-3) and (-3,-3,0)

I think you are able to continue right now. Just do what a answered in the first post

marlon

5. Sep 14, 2004

### marlon

for example a would be (7,2,-3) or S =7x + 2y - 3z

regards
marlon

6. Sep 14, 2004

### HallsofIvy

Staff Emeritus
It is more common to use i, j, and k where you are using x, y, z but I think I understand what you mean:

C = (4x-1y-3z) is what I would call 4i-1j-3k or, more simply just <4, -1, -3> where the basis vectors are understood to be in the x, y, z, directions.

So, C= <4, -1, -3> and D= <2, -3, -5>. Surely your text bood mentions early that vectors written in "component" form can be added and subtracted just by working with each "component" separately. S= C- D is simply <4- 2, -1-(-3),-3-(-5)> =
<2, 2, 2>.

The length of a vector comes from the Pythagorean theorem: $\sqrt{2^2+ 2^2+ 2^2}= \sqrt{12}= 2\sqrt{3}$.

A "unitary" vector (I would say "unit" vector) in a given direction is a vector of length 1 in that direction. In particular, for any vector S, to find a unit vector in the same direction, just divide S by its length: S/|S| which is, again, done component by component. In this problem, since S= <2, 2, 2> and |S|= $2\sqrt{3}$, The unit vector in the direction of S would be $\frac{<2, 2, 2>}{2\sqrt{3}}$.
That is, of course, $\{\frac{1}{\sqrt{3},\frac{1}{\sqrt{3},\frac{1}{\sqrt{3}>$. It is in the same direction as S simply because it has all components the same and its length is $\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}= 1$.

7. Sep 14, 2004

### marlon

Isn't the second vector 2x-3y-5x??? Or is it 2x-3y-5z. In that case my previous answer is not true. Then you would have (4,-1,-3) - (2,-3,-5)=(2,2,2)

marlon