I am looking for find solutions following.
(A x B) x A = 0, B.B=4, A.B=-6, B.C=6
(hint, (A x B) x C = (A.C)B-(A.B)C )
1. A.A=?
2. A.C=?
3. A x B=?
4. A.(B x C)=?
I am looking for find solutions following.
(A x B) x A = 0, B.B=4, A.B=-6, B.C=6
(hint, (A x B) x C = (A.C)B-(A.B)C )
1. A.A=?
2. A.C=?
3. A x B=?
4. A.(B x C)=?
Hi, tykim. There are probably many ways to get what you're after, but I will send you a couple thoughts I had while looking at the problems. Here's a couple of hints to get you going:
1. In the hint replace C with A on both sides of the equation. The left hand side is now (AxB)xA, which we know something about from the given information. The right hand side will have A.A in front of B and A.B in front of A. We know something about A.B so we can use this piece of information here. At this point we should be able to write (A.A)B=-6A. Take the dot product with B on the left and right sides and use what we know about B.B and A.B. I think we are now in a position to determine what A.A is.
2. For part 2. I thought about starting with 0=(AxA)xC <------ Think about this for a little and if it's unclear why this is the case just ask and I'll fill you in. Using 0=(AxA)xC we are basically starting from a similar position that we did in problem 1. Now use the hint like we did in part 1. to expand the cross product of 0=(AxA)xC in terms of the dot product and proceed in a way very much like we did above.
I'll stop for now. Exercises like the ones you've listed are typically solved using little tricks like we did above. If anything is unclear let me know and I can provide more details. Good luck!
Because I am seeking to avoid long calculations as much as possible, I took another look at problem 1 and think I have a more streamlined approach to solving it. The reasoning goes something like this:
Alternative Method for Exercise #1
The condition $\displaystyle 0=(A\times B)\times A$ (along with the other given conditions) actually tells us that $\displaystyle A=kB$, where $\displaystyle k$ is a constant to be determined. <------ Think about this and if it needs further explanation let me know.
Now take the dot product on both sides of $\displaystyle A=kB$ with $\displaystyle B$ and use the given information.
At this point you should know what $\displaystyle k$ is.
Now take $\displaystyle A=kB$ (where you should be using the value for $\displaystyle k$ you determined in the previous step) and dot product both sides with $\displaystyle A$ and use the given information. At this point we should have determined what $\displaystyle A\cdot A$ is.
Let me know if this method needs further explanation. Good luck!
Thanks for the answer.
I guess you are very good at math.
Since I am in bio-field, little chance to meet vector peoblems. Even simple ones for math guy, I feel difficulty to solve. So can you show me more information?
TYKim
I will provide more details for part a below. If any of the steps are unclear please let me know.
Detailed Explanation for Exercise #1
1. The information $\displaystyle B\cdot B=4$ tells us that $\displaystyle B$ is NOT the zero vector (because if $\displaystyle B$ was the zero vector, then $\displaystyle B\cdot B=0$).
2. Similarly, the information $\displaystyle A\cdot B=-6$ tells us that $\displaystyle A$ is NOT the zero vector (because if $\displaystyle A$ was the zero vector, then $\displaystyle A\cdot B=0$).
Now the next part of reasoning may seem a little tricky, but if you think about it enough and ask questions about anything that confuses you, I'm sure you will understand.
3. We now look at the condition $\displaystyle 0=(A\times B)\times A$. We must ask ourselves, how can the cross product $\displaystyle 0=(A\times B)\times A$ be zero? (At this point you should think about this for yourself before moving on).
There are three possible ways that $\displaystyle 0=(A\times B)\times A$:
Possibility 1: $\displaystyle A$ is the zero vector. However, we showed in step 2 that $\displaystyle A$ is NOT the zero vector
Possibility 2: $\displaystyle (A\times B)$ is the zero vector.
Possibility 3: $\displaystyle (A\times B)$ is parallel to $\displaystyle A$. But what do we know about the cross product $\displaystyle A\times B$? We know that it should be PERPENDICULAR to $\displaystyle A$, so it can't be parallel to $\displaystyle A$.
So the only possibility that can happen is Possibility 2: $\displaystyle 0=A\times B$.
4. We now know that $\displaystyle 0=A\times B$. Just like we did in step 3, we ask ourselves, how can $\displaystyle 0=A\times B$? Again we have three possibilities:Since Possibilities 1 & 2 don't happen, Possibility 3 is the truth. Since $\displaystyle A$ and $\displaystyle B$ are parallel, that tells us that $\displaystyle A=kB$, where $\displaystyle k$ is a constant we need to determine.
Possibility 1: $\displaystyle A$ is the zero vector. However, we showed in step 2 that $\displaystyle A$ is NOT the zero vector.
Possibility 2: $\displaystyle B$ is the zero vector. However, we showed in step 1 that $\displaystyle B$ is NOT the zero vector.
Possibility 3: $\displaystyle A$ and $\displaystyle B$ are parallel vectors.
5. We know that $\displaystyle A=kB$. Our goal in this step is to determine the constant $\displaystyle k$. By taking the dot product on both sides of $\displaystyle A=kB$ with $\displaystyle B$ we getFrom the given information we know $\displaystyle A\cdot B=-6$ and$\displaystyle B\cdot B=4$. Plug these numbers in the above equation to get
$\displaystyle A\cdot B=k B\cdot B$.6. We now know $\displaystyle k=-3/2$, so $\displaystyle A=-\frac{3}{2}B$. Now take the dot product on both sides of $\displaystyle A=-\frac{3}{2}B$ with A to get
$\displaystyle -6=4k\Longrightarrow k=-3/2$.From the given information we know $\displaystyle B\cdot A=-6$. Plugging this in the last equation, we get
$\displaystyle A\cdot A=-\frac{3}{2}B\cdot A$.
$\displaystyle A\cdot A=-\frac{3}{2}(-6)=9$
It may seem complicated, but if you're patient with yourself and ask questions about the parts that don't make sense you'll get it. Good luck!
Thanks GJA,
You make me inspired so much and very helpful.
Before posted the problems on the forum, I have tried to solve it other way. Since I could not confirm the below, then posted it to your forum.
Can you go over it below? Is it reasonable with problem 1 and 2?
B∙B=|B||B|cosθ=4
A∙B=|A||B|cosθ=-6, then |A|=(-6)/(|B|cosθ)
B∙C=|B||C|cosθ=6, then |C|=6/(|B|cosθ)
Problem 1. A∙A=|A||A|cosθ=(-6)/(|B|cosθ )×(-6)/(|B|cosθ )×cosθ=36/(|B||B|cosθ)=36/(B∙B)=36/4=9
Problem 2. A∙C=|A||C|cosθ=(-6)/(|B|cosθ )×6/(|B|cosθ )×cosθ=(-36)/(|B||B|cosθ)=(-36)/(B∙B)=(-36)/4=-9
It looks like a good attempt; I like the method you've come up with.
I think there is some confusion when it comes to $\displaystyle \cos\theta$. It appears you're using $\displaystyle \cos\theta$ in two different ways in your equations. What we have to remember is that $\displaystyle \theta$ is the angle between the two vectors we are considering. So when we look at $\displaystyle A\cdot A$ we know $\displaystyle \theta=0$, because the angle between $\displaystyle A$ and $\displaystyle A$ is zero. But when we look at $\displaystyle A\cdot B$, $\displaystyle \theta$ is the angle between $\displaystyle A$ and $\displaystyle B$, and we don't know what this angle is to start with.
If we combine your work with what I posted previously we will have a valid argument. In what follows, $\displaystyle \theta_{AB}$ to denote the angle between vectors $\displaystyle A$ and $\displaystyle B$.
Like you said, we know
1. $\displaystyle |A|=-\frac{6}{|B|\cos\theta_{AB}}$
Notice that instead of using just $\displaystyle \theta$ we're using $\displaystyle \theta_{AB}$ above, because we want to remind ourselves that the angle we must use in this case is the one between the vectors $\displaystyle A$ and $\displaystyle B$
.Now let's look at $\displaystyle A\cdot A$. Like you said, we know
2. $\displaystyle A\cdot A=|A||A|\cos\theta_{AA}$
Now $\displaystyle \theta_{AA}=0$, because the angle between $\displaystyle A$ and $\displaystyle A$ is zero.
Hence, $\displaystyle \cos\theta_{AA}=1$. Equation 2 then becomes
3. $\displaystyle A\cdot A = |A||A|=|A|^{2}$
Using equation 1 in equation 3 we get
4.$\displaystyle
\begin{align*}
A\cdot A &=\left (-\frac{6}{|B|\cos\theta_{AB}}\right)^{2}\\
&=\frac{36}{|B|^{2}\cos^{2}\theta_{AB}}
\end{align*}$
At this point we need combine your work with what was done in the previous post.
Remember from the previous post that we determined the vectors $\displaystyle A$ and $\displaystyle B$ must be parallel. This means $\displaystyle \theta_{AB}=0$ or $\displaystyle \theta_{AB}=180$. This means either $\displaystyle \cos\theta_{AB}=1$ or $\displaystyle \cos\theta_{AB}=-1$. But notice that in the last line equation 4 we are squaring $\displaystyle \cos\theta_{AB}$. So it doesn't matter if $\displaystyle \cos\theta_{AB}$ is 1 or -1, because after we square we get 1 in both cases.
Equation 4 is now
5. $\displaystyle A\cdot A=\frac{36}{|B|^{2}}$
Now using the given information
$\displaystyle 4=B\cdot B=|B||B|\cos\theta_{BB}=|B||B|=|B|^{2}$
Above we used the fact that $\displaystyle \theta_{BB}=0$ because the angle between the vector $\displaystyle B$ and itself is 0, so $\displaystyle \cos\theta_{BB}=1$. Plugging $\displaystyle |B|^{2}=4$ in equation 5 we get
$\displaystyle A\cdot A=\frac{36}{4}=9$.
So you can see that you had a nice method for doing the problem. We just needed to be a little bit careful to keep track of what $\displaystyle \theta$ means each time we consider a different dot product. Once you combine your method with what we knew from the previous post your argument is complete. Nice job! If you have any other questions feel free to ask.