# Vectors

• January 13th 2007, 09:52 PM
gretchen
Vectors
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.
• January 14th 2007, 12:07 AM
earboth
Quote:

Originally Posted by gretchen
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.

Hello,

5a=(15, 25). Thus

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

EB
• January 14th 2007, 12:17 AM
CaptainBlack
Quote:

Originally Posted by gretchen
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.

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
• January 14th 2007, 03:18 AM
gretchen
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.
• January 14th 2007, 03:57 AM
topsquark
Quote:

Originally Posted by gretchen
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.

5a - 3b = 5(3i - 2j) - 3(4i + 4j) = 15i - 10j - 12i -12j = (15i - 12i) + (-10j -12j) = 3i - 22j

-Dan
• January 14th 2007, 04:05 PM
gretchen
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
• January 14th 2007, 04:10 PM
ThePerfectHacker
Quote:

Originally Posted by gretchen
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

Just double it,
6i-14j

Doubling vector coordinates keeps it in the same direction and twice as long.
• January 14th 2007, 04:31 PM
gretchen

Find the dot product of the vectors i-3j and 4i+2j.
• January 14th 2007, 07:11 PM
ThePerfectHacker
Quote:

Originally Posted by gretchen

Find the dot product of the vectors i-3j and 4i+2j.

Multiply the components.
Meaning the i with the i and j with the j.

Thus,
$(1)(4)+(-3)(2)$
Is the dot product.
• January 14th 2007, 09:01 PM
gretchen
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.
• January 14th 2007, 09:59 PM
earboth
Quote:

Originally Posted by gretchen
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.

Hello, Gretchen,

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

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

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

Use this formula with your vectors:

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

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.
• January 15th 2007, 06:18 AM
ThePerfectHacker
Quote:

Originally Posted by gretchen
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.

The formula is,
$\bold{u}\cdot \bold{v} = ||\bold{u} ||\cdot ||\bold{v}|| \cos \theta$
Where,
$\bold{u}\cdot \bold{v}$---> Dot product.
$||\bold{u}||\cdot ||\bold{v}||$---> Product of their norms.
And,
$\cos \theta$ is the angle between them.