SOLVEDVectors.

Altami

Hey everybody I know I have been posting a lot and I am sorry, I just really want to be sure that I am doing my H.W. right and getting the right answers.

Vector A = <1,-4> and vector B = <5,7>

a) Find the vector C=2A-B

For C I got <-3,1>

b) Find the magnitude of C

For this I got /square root 10/

c) Find 'Theta' the direction of C such that 0degrees is 'less than or equal to' "theta" < 360 degrees.

d) Find the angle between the vectors A and B...

The first two I just need verification but the last two I couldn't get

If anybody can help me...thanks.

Prove It

MHF Helper
Your $$\displaystyle \mathbf{C}$$ is wrong.

$$\displaystyle \mathbf{C} = 2\mathbf{A} - \mathbf{B}$$

$$\displaystyle = 2(1, -4) - (5, 7)$$

$$\displaystyle = (2, -8) - (5, 7)$$

$$\displaystyle = (2-5, -8-7)$$

$$\displaystyle = (-3, -15)$$.

Now fix up everything that follows.

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Altami

Ok, so I ended up getting

Like you said <-3,-11> for C and the magnitude I got /square root 130/

I think the magnitude is right?

But again I don't really know how to do the other two?

earboth

MHF Hall of Honor
Your $$\displaystyle \mathbf{C}$$ is wrong.

$$\displaystyle \mathbf{C} = 2\mathbf{A} - \mathbf{B}$$

$$\displaystyle = 2(1, -4) - (5, 7)$$

$$\displaystyle = (2, -4) - (5, 7)$$ <=== unfortunately you made a small mistake here

$$\displaystyle = (2-5, -4-7)$$

$$\displaystyle = (-3, -11)$$.

Now fix up everything that follows.
$$\displaystyle = 2(1, -4) - (5, 7)$$

$$\displaystyle = (2, -8) - (5, 7) = (-3, -15)$$

and $$\displaystyle |(-3, -15)| = \sqrt{234}$$

Prove It

Prove It

MHF Helper
$$\displaystyle = 2(1, -4) - (5, 7)$$

$$\displaystyle = (2, -8) - (5, 7) = (-3, -15)$$

and $$\displaystyle |(-3, -15)| = \sqrt{234}$$
Yes I did, Gah! Thanks. Editing now.

Altami

Jesus this is a great forum, you guys really help each other out. I thank you guys for helping me as well. I was actually wondering about why only the "1" had gotten multiplied by two and not the "-4". Anyways thanks guys for helping me!

But again, the last two I don't really know how even start...if you guys could give me any tips?

By last two I mean "C and D"

earboth

MHF Hall of Honor
Jesus this is a great forum, you guys really help each other out. I thank you guys for helping me as well. I was actually wondering about why only the "1" had gotten multiplied by two and not the "-4". Anyways thanks guys for helping me!

But again, the last two I don't really know how even start...if you guys could give me any tips?

By last two I mean "C and D"
To be honest: I don't understand the question C)

To D: Use the definition of the dot product:

$$\displaystyle \vec a \cdot \vec b = |\vec a| \cdot |\vec b| \cdot \cos(\angle(\vec a, \vec b)~\implies~\cos(\angle(\vec a, \vec b) = \dfrac{\vec a \cdot \vec b}{|\vec a| \cdot |\vec b|}$$

Altami

I don't now how you guys use that perfect numerical abbreviations but anyways.

For D). I got 77.2degrees....

Would that be right?

Altami

Ok I spent like 10 minutes trying to figure out how to use Latex...anyways question C supposed to look like this

Find $$\displaystyle \theta$$ the direction of C such that $$\displaystyle 0\deg \leq \theta < 360\deg$$

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earboth

MHF Hall of Honor
I don't now how you guys use that perfect numerical abbreviations but anyways.
Have a look here: http://www.mathhelpforum.com/math-help/latex-help/266-latex-tutorial.html

For D). I got 77.2degrees....

Would that be right?
I don't know how you got this result but here is what I've done:

$$\displaystyle \cos(\angle(\vec a, \vec b)) = \dfrac{(1, -4) \cdot (5, 7)}{\sqrt{17} \cdot \sqrt{74}} = \dfrac{-23}{\sqrt{1258}}$$

which yields $$\displaystyle \angle(\vec a, \vec b) \approx 130.426^\circ$$

If you use absolute values you'll get an angle of $$\displaystyle 49.574^\circ$$

Altami