# Cross and dot product vectors

• Jul 5th 2010, 12:44 AM
arze
Cross and dot product vectors
axb=2axc

b-2c=$\displaystyle \lambda$a

given that |a|=|c|=1 and |b|=4
and the angle between b and c is $\displaystyle \arccos \frac{1}{4}$
show that $\displaystyle \lambda =+4 , -4$
For each of these cases find the cosine of the angle between a and c.

I don't know where to begin
• Jul 5th 2010, 02:50 AM
Swlabr
Quote:

Originally Posted by arze
axb=2axc

b-2c=$\displaystyle \lambda$a

given that |a|=|c|=1 and |b|=4
and the angle between b and c is $\displaystyle \arccos \frac{1}{4}$
show that $\displaystyle \lambda =+4 , -4$
For each of these cases find the cosine of the angle between a and c.

I don't know where to begin

I suppose your first step would be to work out what to do with the arccos. So, where should that go? What does this tell you?

Secondly, you should work out that $\displaystyle a \times (b-2c) = 0$, and so that means the second proposition makes sense - the vectors are in the same direction. That is to say, they only differ by a scalar.

What you now want to do is prove that $\displaystyle |b-2c|=4$ (why?). So, work out what $\displaystyle \sqrt{(b-2c).(b-2c)}$ is, and you will be done (why?)! (You will need the information you gleamed from my first line here).

Does that make sense?
• Jul 5th 2010, 03:11 AM
arze
I'm not sure I exactly understand. Cross products have the sin value multiplied by the magnitudes, right? So that's confusing me. I did work out $\displaystyle a \times (b-2c) = 0$ and so $\displaystyle (b-2c)=\lambda a$ and $\displaystyle \lambda$ is a scalar. Ok, I understand why $\displaystyle |b-2c|=4$, since $\displaystyle \lambda=\pm 4$. I don't get the last one about $\displaystyle \sqrt{(b-2c).(b-2c)}$.
Thanks!
• Jul 5th 2010, 03:17 AM
Swlabr
Quote:

Originally Posted by arze
I'm not sure I exactly understand. Cross products have the sin value multiplied by the magnitudes, right? So that's confusing me. I did work out $\displaystyle a \times (b-2c) = 0$ and so $\displaystyle (b-2c)=\lambda a$ and $\displaystyle \lambda$ is a scalar. Ok, I understand why $\displaystyle |b-2c|=4$, since $\displaystyle \lambda=\pm 4$. I don't get the last one about $\displaystyle \sqrt{(b-2c).(b-2c)}$.
Thanks!

You do not know that $\displaystyle |b-2c| = 4$, this is what you have to prove!

To do this, you need to remember that $\displaystyle |v| = \sqrt{v.v}$. To work out what this is, you need to know some dot products. The arccos that you have been given will help you in this.
• Jul 5th 2010, 03:55 PM
arze
$\displaystyle |b-2c|=\sqrt{b.b-4b.c+4c.c}=\sqrt{5-4|4||1|(\frac{1}{4})}=\sqrt{1}=\pm 1$
I'm still doing something wrong.
• Jul 5th 2010, 04:01 PM
Plato
Quote:

Originally Posted by arze
$\displaystyle |b-2c|=\sqrt{b.b-4b.c+4c.c}=\sqrt{5-4|4||1|(\frac{1}{4})}=\sqrt{1}=\pm 1$
I'm still doing something wrong.

I think this is what you are missing.
$\displaystyle \left( {b - 2c} \right) \cdot \left( {b - 2c} \right) = b \cdot b - 4b \cdot c + 4c \cdot c = \left\| b \right\|^2 - 4b \cdot c + 4\left\| c \right\|^2$
• Jul 5th 2010, 04:04 PM
arze
Quote:

Originally Posted by Plato
I think this is what you are missing.
$\displaystyle \left( {b - 2c} \right) \cdot \left( {b - 2c} \right) = b \cdot b - 4b \cdot c + c \cdot c = \left\| b \right\|^2 - 4b \cdot c + 4\left\| c \right\|^2$

What about the number 4 from the 2c?
Thanks alot!