# Magnitude, Unit vectors,unknown variables

1. Mar 4, 2014

### concon

1. The problem statement, all variables and given/known data
a and b are vectors in R^3 s.t. a=(1,7,-4) and b= -3j-4k

1. Find ||3a-3b|| (magnitude of 3a-3b)
2. Find unit vector u in direction of 3a-3b, write answer in form (u1,u2,u3)
3. Find vector of length 8 in direction of a (write answer in form "-")

2. Relevant equations
||3a-3b|| = 3||a-b|| = 3*sqrt((a1-b1)^2 +.......(a3-b3)^2)
3 is positve scalar and can be factored out I believe

U = X/ ||X|| , X ≠ 0

Length = sqrt(x^2 + y^2) (might be wrong, I am confused on finding length in same dir)

3. The attempt at a solution

Starting solving 1.
3||a-b|| = 3 sqrt((1+3j+4k)^2 + (7+3j+4k)^2 + (-4+3j+4k)^2)
I don't know what j and k represent

2. I would need to know ||3a-3b|| first right?
then do X/||X||?

3. a= (1,7,-4) and L=8
L is in same direction of a so
8 = sqrt((u1-1)^2 + (u2-7)^2 + (u3 +4)^2) -> or maybe the a vals and u's are switched i don't know

2. Mar 4, 2014

### micromass

Staff Emeritus
Not sure where this comes from.

The formula is that if we are given $a = (a_1,a_2,a_3)$ and $b = (b_1, b_2, b_3)$, then

$$\|a-b\|= \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2 + (a_3 - b_3)^2}$$

Here you are given $a= (1,7,-4)$ and $b=(0,-3,-4)$.

3. Mar 4, 2014

### concon

Wow wait a minute, how do you know the values for b as (0,-3, -4)?
What do j and k represent?

Also, do you have any input regarding the other two questions I posted w/in this post?

4. Mar 4, 2014

### micromass

Staff Emeritus
The notation $-3j -4k$ is just another notation for $(0,-3,-4)$.

The vector $i$ actually equals $(1,0,0)$. The vector $j=(0,1,0)$ and $k=(0,0,1)$. So

$$-3-4k = -3(0,1,0) -4 (0,0,1) = (0,-3,0) + (0,0,-4) = (0,-3,-4)$$

Sure, but let's do (a) first.

5. Mar 4, 2014

### concon

Okay gotcha, thanks! My prof did not cover that this lecture so I was unclear.

Okay so with that in mind ||a-b|| = sqrt(101) right?
and since 3 is a postive number it can be factored out thus,

3*sqrt(101) = ||3a-3b|| right?

6. Mar 4, 2014

### micromass

Staff Emeritus
Yes, that's correct.

So, can you use this to solve (b) now?

7. Mar 4, 2014

### concon

I think so but is my equation correct? Which values am I using

8. Mar 4, 2014

### concon

actually scratch that I got the answer. I got:
U = (1/sqrt 101, 10/sqrt 101, 0)

How do I do the last part about length?

9. Mar 5, 2014

### micromass

Staff Emeritus
Given your vector $a$, then if $\lambda>0$ is a real number, then $\lambda a$ lies in the same direction of $a$. So your answer will be $\lambda a$. Your only job is to figure out what $\lambda$ is exactly.

You want $\|\lambda a\| = 8$. Can you figure out what $\lambda$ is now?

10. Mar 5, 2014

### concon

yep got the answer as (8/sqrt 66 , 56/sqrt 66 , -32/sqrt 66)

11. Mar 5, 2014

### micromass

Staff Emeritus
Right!