I hope someone can help me with this, I have never posted here before.

Given vectors a=(3,5) and b=(2,3). Find 5a-b.

Thanks if you can help me.

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- Jan 13th 2007, 09:52 PMgretchenVectors
I hope someone can help me with this, I have never posted here before.

Given vectors a=(3,5) and b=(2,3). Find 5a-b.

Thanks if you can help me. - Jan 14th 2007, 12:07 AMearboth
- Jan 14th 2007, 12:17 AMCaptainBlack
Multiplying a vector by a scalar (such as 5a here) multiplies each component

of the vector by the scalar (so 5a=(5*3, 5*5)=(15, 25)). Adding or

subtracting two vectors gives a vector whoese components are the sum of

difference of the corresponding components of the two vectors.

So:

5a-b=5(3, 5) - (2, 3) = (15, 25) - (2, 3) = (15-2, 25-3) = (13, 22)

RonL - Jan 14th 2007, 03:18 AMgretchen
Thanks for the help, I see what I had to do now. What about going at it from a different perspective?

Given vectors a=3i-2j and b=4i+4j. Find 5a-3b. - Jan 14th 2007, 03:57 AMtopsquark
- Jan 14th 2007, 04:05 PMgretchen
Thanks guys for your help. I have another one that is confusing me:

Find the vector that has the same direction as 3i-7j and twice the magnitude.

Thanks,

Gretch - Jan 14th 2007, 04:10 PMThePerfectHacker
- Jan 14th 2007, 04:31 PMgretchen
What about when asking for a dot product:

Find the dot product of the vectors i-3j and 4i+2j. - Jan 14th 2007, 07:11 PMThePerfectHacker
- Jan 14th 2007, 09:01 PMgretchen
Thanks, I will work the rest like that. Is there any trick to coming up with angles from two vectors?

Find the angle between the two vectors (5,2) and (-2, 5).

Thanks again. - Jan 14th 2007, 09:59 PMearboth
Hello, Gretchen,

there isn't any trick to calculate the angle - only a simple formula:

Let $\displaystyle \alpha$ be the angle between the vectors $\displaystyle \vec a$ and $\displaystyle \vec b$ then you get the angle by:

$\displaystyle \cos(\alpha)=\frac{\vec a \cdot \vec b}{|\vec a| \cdot |\vec b|}$

Use this formula with your vectors:

$\displaystyle \cos(\alpha)=\frac{(5,2) (-2 ,5)}{\sqrt{5^2+2^2} \cdot \sqrt{(-2)^2+5^2}}=\frac{0}{29}$. Thus $\displaystyle \alpha = 90^\circ$

By the way: This result shows: Two vectors are perpendicular if their dot-product equals zero.

EB

PS: Please do us a favour and start a new thread if you have a new problem to do. Otherwise you risk that nobody will notice that you ask for some help again. - Jan 15th 2007, 06:18 AMThePerfectHacker
The formula is,

$\displaystyle \bold{u}\cdot \bold{v} = ||\bold{u} ||\cdot ||\bold{v}|| \cos \theta $

Where,

$\displaystyle \bold{u}\cdot \bold{v}$---> Dot product.

$\displaystyle ||\bold{u}||\cdot ||\bold{v}||$---> Product of their norms.

And,

$\displaystyle \cos \theta$ is the angle between them.