Find a Vector given a length

**craig** · October 19th 2009, 11:45 AM

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

**tonio** · October 19th 2009, 12:50 PM

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

**craig** · October 19th 2009, 01:45 PM

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?

**tonio** · October 19th 2009, 10:58 PM

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

**HallsofIvy** · October 20th 2009, 05:38 AM

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).

**craig** · October 20th 2009, 07:36 AM

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

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