# Homework Help: Finding vectors and magnitude

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1. Jun 9, 2015

### mmont012

1. The problem statement, all variables and given/known data

Find a vector in the direction opposite to <-4,1,2>, that has a magnitude of 3.

2. Relevant equations
I think that I did the first part of the problem correctly:
<-4,1,2>
magnitude= sqrt[ (-4)^2+1^2+2^2 ]
= sqrt(16+1+4)
= sqrt(21)
(-4/sqrt(21), 1/sqrt(21), 2/sqrt(21) )
to get the opposite direction, I would just change the signs
(4/sqrt(21), -1/sqrt(21), -2/sqrt(21) )

But I am confused at where to go from here. What does it mean "that has a magnitude of 3"? I may have approached this problem incorrectly. Any help would be great. Thank you for stopping by.

2. Jun 9, 2015

### Zondrina

You have found a normal vector. The magnitude of a normal vector is $1$.

Perhaps the answer is more trivial than it first appears, just think:

$$\sqrt{a^2 + b^2 + c^2} = 3$$

Select a vector $<a, b, c>$ such that the above is satisfied. Multiplying the original vector by $-1$ will help. In fact, one is the loneliest one, and the only one you'll ever need for this problem.

3. Jun 9, 2015

### Staff: Mentor

As Zondrina has pointed out, you have correctly determined a unit vector in the desired direction. Now, what do you need to multiply this unit vector by to produce a vector having the same direction, but with a magnitude of 3?

Chet

4. Jun 9, 2015

### mmont012

Ah, okay. I think that I was making this problem more difficult than it actual is. Thank you so much for the help!

5. Jun 10, 2015

### Staff: Mentor

Removed "Calc III" from thread title -- this is really a precalc type of problem. Also moved thread to the precalc forum section.

6. Jun 23, 2015

### mmont012

How is this a pre-calc problem if I'm in calc 3 and this is the first time that I've dealt with this? Did my university just not cover this? Or am I behind?

7. Jun 23, 2015

### Staff: Mentor

Vectors in R2 and R3 are often covered in precalculus with trig courses.

8. Jun 23, 2015

### SteamKing

Staff Emeritus
No limits, no derivatives, no integrals = no calculus.

Your calc 3 course may simply be providing a refresher on basic vector math before jumping into vector calculus.
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