# Thread: Find a Vector given a length

1. ## Find a Vector given a length

Hi, it's been a while since I've attempted anything to do with vectors and I'm a little rusty at the moment.

The question is:

Fine a vector length $2\sqrt{6}$ anti-parallel to $d = (-1,2,-1)$.

I thought that maybe the vector product would be the way forward but no idea where to start.

Any help would be greatly appreciated.

Thanks

Craig

2. Originally Posted by craig
Hi, it's been a while since I've attempted anything to do with vectors and I'm a little rusty at the moment.

The question is:

Fine a vector length $2\sqrt{6}$ anti-parallel to [tex]d = (-1,2,-1)[/MATH.

I thought that maybe the vector product would be the way forward but no idea where to start.

Any help would be greatly appreciated.

Thanks

Craig

What about $2\left(1,-2,1\right)=\left(2,-4,2\right)$ ?

Tonio

3. Originally Posted by tonio
What about $2\left(1,-2,1\right)=\left(2,-4,2\right)$ ?

Tonio
Hi Tonio thank's for the reply.

How did you come up with this answer?

4. Originally Posted by craig
Hi Tonio thank's for the reply.

How did you come up with this answer?

Antiparallel ==> it has to "go the other way" so multiply by $-1$ all its entries.
Now, the vector given and the new one both have norm $=\sqrt{6}$ so in order to get norm equal to $2\sqrt{6}$ simply multiply the vector by 2.

Tonio

5. Originally Posted by craig
Hi, it's been a while since I've attempted anything to do with vectors and I'm a little rusty at the moment.

The question is:

Fine a vector length $2\sqrt{6}$ anti-parallel to $d = (-1,2,-1)$.

I thought that maybe the vector product would be the way forward but no idea where to start.

Any help would be greatly appreciated.

Thanks

Craig
Any vector parallel to (-1, 2, -1) must be of the form a(-1, 2, -1)= (-a, 2a, -a). Any vector anti-parallel to (-1, 2, -1) (in the opposite direction) must be of that form with a negative- or we can write it -a(-1, 2, -1)= (a, -2a, a) with the requirement that a be positive.

Now, the length of (a, -2a, a) is $\sqrt{a^2+ (-2a)^2+ a^2}= \sqrt{6a^2}= a\sqrt{6}$. In order that the vector have length $2\sqrt{6}$, we must have $a\sqrt{6}= 2\sqrt{6}$ or a= 2. (a, -2a, a)= (2, -4, 2).

6. Thanks for both replys, all makes sense now. It's been a very long time since I've attempted anything to do with vectors, got quite a bit of catching up to do