# Simple vectors

• June 19th 2008, 10:53 AM
Mathstud28
Simple vectors
My friend asked me to help him with his homework, and I think I might have forgotten how to do it (Thinking)

They were all like

"Find a vector with given magnitude and in the same direction as the given vector"

So I think what you would do is this

Given the position vector

$$

and the new vector in the same position must be of magnitude $d$

Would I just do this

the new vector must be of the form

$\varpi$

where $\varpi$ is an abritrary constant, so now I just have to find that constant such that the new magnitude is

Would I do this by solving

$d=\sqrt{(a\varpi)^2+(b\varpi)^2+(c\varpi)^2}$

For $\varpi$?

And then the answer will be of the form $|\varpi=|\text{whatever}
\Rightarrow\varpi=\pm\text{whatever}$

So should I then take the $\varpi$ that when plotted goes in the same direction, because won't one of them be parallel but in the opposite direction?
• June 19th 2008, 10:56 AM
Moo
Hello,

Quote:

Originally Posted by Mathstud28
My friend asked me to help him with his homework, and I think I might have forgotten how to do it (Thinking)

They were all like

"Find a vector with given magnitude and in the same direction as the given vector"

So I think what you would do is this

Given the position vector

$$

and the new vector in the same position must be of magnitude $d$

Would I just do this

the new vector must be of the form

$\varpi$

where $\varpi$ is an abritrary constant, so now I just have to find that constant such that the new magnitude is

Would I do this by solving

$d=\sqrt{(a\varpi)^2+(b\varpi)^2+(c\varpi)^2}$

For $\varpi$?

Same direction + same sense would yield $\varpi>0$

So $d=\varpi \sqrt{a^2+b^2+c^2}$

Your method looks ok to me..
• June 19th 2008, 10:58 AM
TheEmptySet
Quote:

Originally Posted by Mathstud28
My friend asked me to help him with his homework, and I think I might have forgotten how to do it (Thinking)

They were all like

"Find a vector with given magnitude and in the same direction as the given vector"

So I think what you would do is this

Given the position vector

$$

and the new vector in the same position must be of magnitude $d$

Would I just do this

the new vector must be of the form

$\varpi$

where $\varpi$ is an abritrary constant, so now I just have to find that constant such that the new magnitude is

Would I do this by solving

$d=\sqrt{(a\varpi)^2+(b\varpi)^2+(c\varpi)^2}$

For $\varpi$?

Just make your vector a unit vector then multiply the unit by the magnitude that you want

$\vec v=$ and you want a vector in the same direction with magnitude M then

$M\frac{\vec v}{|\vec v|}=\frac{M}{\sqrt{a^2+b^2+c^2}}\vec v$
• June 19th 2008, 10:59 AM
Mathstud28
Quote:

Originally Posted by Moo
Hello,

Same direction + same sense would yield $\varpi>0$

So $d=\varpi \sqrt{a^2+b^2+c^2}$

Your method looks ok to me..

So how would I discern between the $|\varpi|$ giving me a plus or minus answer, can I tell which I want without graphing them?

Thanks Moo
• June 19th 2008, 11:00 AM
Moo
Quote:

Originally Posted by Mathstud28
So how would I discern between the $|\varpi|$ giving me a plus or minus answer, can I tell which I want without graphing them?

Thanks Moo

If it's stated that the direction and the sense (for example, to the top) are the same as the original vector (which would be directed to the top too), you can say that $\varpi>0$. Otherwise, I'm afraid we would need more information.

But I'm not sure I understood correctly the question that your friend asked, what exactly is given, etc (Tongueout)
• June 19th 2008, 11:01 AM
Mathstud28
Quote:

Originally Posted by TheEmptySet
Just make your vector a unit vector then multiply the unit by the magnitude that you want

$\vec v=$ and you want a vector in the same direction with magnitude M then

$M\frac{\vec v}{|\vec v|}=\frac{M}{\sqrt{a^2+b^2+c^2}}\vec v$

...good point(Smirk)....I guess this is payback for the odd functions thing (Wink)
• June 19th 2008, 11:07 AM
TheEmptySet
Quote:

Originally Posted by Mathstud28
...good point(Smirk)....I guess this is payback for the odd functions thing (Wink)

No thanks for that! (Clapping)
• June 19th 2008, 11:11 AM
Mathstud28
Quote:

Originally Posted by TheEmptySet
No thanks for that! (Clapping)

Haha, I was just kidding, thanks for giving me an easier way to do this thing above

And thank you Moo for validating my semi-correct method.