Simplifying expression with vectors

[olgerm]

Homework Statement

To simplify:

1. $\mid \vec b -\frac{\vec a*\vec b}{\vec a*\vec c} * \vec c \mid$
2. $\mid \vec b -\frac{(\vec a*\vec c)*(\vec b*\vec b)-(\vec a*\vec b)*(\vec b*\vec c)}{(\vec a*\vec b)*(\vec c*\vec c)-(\vec a*\vec c)*(\vec b*\vec c)} * \vec c \mid$

Vectors may be any dimensional!

Homework Equations

$\vec a*\vec b=\sum(a_i*b_i)$
$\mid \vec b \mid=\sqrt{\sum(b_i^2)}$

The Attempt at a Solution

1. $\sqrt{(\vec b*\vec b)*(\vec a*\vec c)^2-2*(\vec a*\vec b)*(\vec b*\vec c)*(\vec a*\vec c)+(\vec c*\vec c)*(\vec a*\vec b)^2}/(\vec a*\vec c)$ is it correct? Can it be more simplified ? Can it be written as one product under square root ?To simplify it I wrote vectors as 3-dimensional vectors and opened braces: $b2^2*a3^2*c3^2+b1^2*a3^2*c3^2+b2^2*a2^2*c3^2+2*b1*a1*b2*a2*c3^2+b1^2*a1^2*c3^2+2*b1^2*a2*a3*c2*c3+2*a1*b2^2*a3*c1*c3+b3^2*a3^2*c2^2+2*b1*a1*b3*a3*c2^2+a2^2*b3^2*c2^2+b1^2*a2^2*c2^2+b1^2*a1^2*c2^2+2*a1*a2*b3^2*c1*c2*+b3^2*a3^2*c1^2+2*b2*a2*b3*a3*c1^2+a1^2*b3^2*c1^2+b2^2*a2^2*c1^2+a1^2*b2^2*c1^2*-2*b1*a1^2*b2*c1*c2-2*b1*a1^2*b3*c1*c3-2*b1*a1*b2*a3*c2*c3-2*b1*a1*a2*b3*c2*c3-2*b1*b2*a2^2*c1*c2-2*b1*b2*a2*a3*c1*c3-2*b1*a2*b3*a3*c1*c2-2*b1*b3*a3^2*c1*c3-2*a1*b2*a2*b3*c1*c3-2*a1*b2*b3*a3*c1*c2-2*b2*a2^2*b3*c2*c3-2*b2*b3*a3^2*c2*c3$ but I can not factorisate it!
2. $\sqrt{(\vec b*\vec b)-2*(\frac{(\vec a*\vec c)*(\vec b*\vec b)-(\vec a*\vec b)*(\vec b*\vec c)}{(\vec a*\vec b)*(\vec c*\vec c)-(\vec a*\vec c)*(\vec b*\vec c)})*(\vec b*\vec c)+(\vec c*\vec c)*(\frac{(\vec a*\vec c)*(\vec b*\vec b)-(\vec a*\vec b)*(\vec b*\vec c)}{(\vec a*\vec b)*(\vec c*\vec c)-(\vec a*\vec c)*(\vec b*\vec c)})^2}$ is it correct? Can it be more simplified ? Can it be written as one product under square root ?

 

[Fightfish]

You should really differentiate your dot products from ordinary multiplications; I'm seeing stars after all the asterisks .
Anyways, it helps greatly if you recognise that |\vec{b}| = \sqrt{(\vec{b}\cdot \vec{b})}, and not rush to expand everything.

 

[haruspex]

Not sure it's a simplification, but are you familiar with the triple vector product?

 

[olgerm]

it helps greatly if you recognise that |\vec{b}| = \sqrt{(\vec{b}\cdot \vec{b})}

I know it, but how it helps?

not rush to expand everything.

What should I do then?

 

[Fightfish]

What should I do then?

Keep the vector forms.

I know it, but how it helps?

As an example, let's say I have a vector \vec{c} = \alpha \vec{a} + \beta\vec{b}, and I want to find its norm. Then its straightforward for me to apply
|\vec{c}|^{2} = \left(\alpha \vec{a} + \beta\vec{b}\right) \cdot \left(\alpha \vec{a} + \beta\vec{b}\right) = \alpha^{2} |\vec{a}|^{2} + 2\alpha\beta \left(\vec{a} \cdot \vec{b}\right) + \beta^{2} |\vec{b}|^{2}

Edit: Okay, I see that's essentially what you have there, although you have a typo in the first answer that threw me off.

 

[haruspex]

Keep the vector forms.

As an example, let's say I have a vector \vec{c} = \alpha \vec{a} + \beta\vec{b}, and I want to find its norm. Then its straightforward for me to apply
|\vec{c}|^{2} = \left(\alpha \vec{a} + \beta\vec{b}\right) \cdot \left(\alpha \vec{a} + \beta\vec{b}\right) = \alpha^{2} |\vec{a}|^{2} + 2\alpha\beta \left(\vec{a} \cdot \vec{b}\right) + \beta^{2} |\vec{b}|^{2}

I'm not seeing anything different here from what olgerm tried in the OP.
@olgerm, there is an interesting relationship between the two expressions (Q1 and Q2). Can you see how to get the second from the first?

 

[Fightfish]

Yeah, my bad, I was somewhat thrown off by a typo he made in his first answer, so I assumed he had messed up somewhere.
If you really want to "simplify" it further, you can ostensibly try to factor out the denominator. That let's you use the triple vector product as Haruspex suggested.

 

[olgerm]

Can you see how to get the second from the first?

$\mid \vec b -B * \vec c \mid$

1. $B=\frac{\vec a*\vec b}{\vec a*\vec c}$
2. $B=\frac{(\vec a*\vec c)*(\vec b*\vec b)-(\vec a*\vec b)*(\vec b*\vec c)}{(\vec a*\vec b)*(\vec c*\vec c)-(\vec a*\vec c)*(\vec b*\vec c)}$

$\sqrt{(\vec b*\vec b)-2*B*(\vec b*\vec c)+B^2*(\vec c*\vec c)}$

 

[olgerm]

Does anybody know, can these be more simplified ,so that simplified equation includes only dot products, or not?

 

[Fredrik]

The dot product symbol is written as \cdot, and the product of two real numbers x and y is written as xy. So you should write $(\vec a\cdot\vec c)(\vec b\cdot\vec b)$ instead of $(\vec a*\vec c)*(\vec b*\vec b)$ for example.

Is there a hint about what the simplified expressions should look like? I mean, it's straightforward to rewrite them as expressions that don't involve sums of vectors, but should those expressions be considered simpler? How do you know if what you get is simple enough?

 

[olgerm]

Homework Statement

To simplify:

1. $\mid \vec b -\frac{\vec a \cdot \vec b}{\vec a \cdot \vec c} \vec c \mid=\sqrt{(\vec b \cdot \vec b)(\vec a \cdot \vec c)^2-2(\vec a \cdot \vec b)(\vec b \cdot \vec c)(\vec a \cdot \vec c)+(\vec c \cdot \vec c)(\vec a \cdot \vec b)^2}/(\vec a \cdot \vec c)$
2. $\mid \vec b -\frac{(\vec a \cdot \vec c)(\vec b \cdot \vec b)-(\vec a \cdot \vec b)(\vec b \cdot \vec c)}{(\vec a \cdot \vec b)(\vec c \cdot \vec c)-(\vec a \cdot \vec c)*(\vec b \cdot \vec c)} \vec c \mid =\sqrt{(\vec b \cdot \vec b)-2(\frac{(\vec a \cdot \vec c)(\vec b \cdot \vec b)-(\vec a \cdot \vec b)(\vec b \cdot \vec c)}{(\vec a \cdot \vec b)(\vec c \cdot \vec c)-(\vec a \cdot \vec c)(\vec b \cdot \vec c)})(\vec b \cdot \vec c)+(\vec c \cdot \vec c)(\frac{(\vec a \cdot \vec c)(\vec b \cdot \vec b)-(\vec a \cdot \vec b)(\vec b \cdot \vec c)}{(\vec a \cdot \vec b)(\vec c \cdot \vec c)-(\vec a \cdot \vec c)(\vec b \cdot \vec c)})^2}$

How do you know if what you get is simple enough?

I want in as simplified form as possible, but i do not know how simple is that!
For example may it be square root of product of 3 dot products?To simplify first one I opened braces and expanded it:
$b2^2*a3^2*c3^2+b1^2*a3^2*c3^2+b2^2*a2^2*c3^2+2*b1*a1*b2*a2*c3^2+b1^2*a1^2*c3^2+2*b1^2*a2*a3*c2*c3+2*a1*b2^2*a3*c1*c3+b3^2*a3^2*c2^2+2*b1*a1*b3*a3*c2^2+a2^2*b3^2*c2^2+b1^2*a2^2*c2^2+b1^2*a1^2*c2^2+2*a1*a2*b3^2*c1*c2*+b3^2*a3^2*c1^2+2*b2*a2*b3*a3*c1^2+a1^2*b3^2*c1^2+b2^2*a2^2*c1^2+a1^2*b2^2*c1^2*-2*b1*a1^2*b2*c1*c2-2*b1*a1^2*b3*c1*c3-2*b1*a1*b2*a3*c2*c3-2*b1*a1*a2*b3*c2*c3-2*b1*b2*a2^2*c1*c2-2*b1*b2*a2*a3*c1*c3-2*b1*a2*b3*a3*c1*c2-2*b1*b3*a3^2*c1*c3-2*a1*b2*a2*b3*c1*c3-2*a1*b2*b3*a3*c1*c2-2*b2*a2^2*b3*c2*c3-2*b2*b3*a3^2*c2*c3$ but I can not factorise it! Can somebody (who for example has Wolfram Mathematica) factorise it for me?

 

[Fredrik]

I think it's a mistake to introduce the components of the vectors into the calculation. Note by the way that you're not allowed to make assumptions about how many of them there are.

What you did in post #8 looks like a good start. The last term under the square root can be simplified a lot. (I have only looked at problem 1). Edit: Uhh, I think I spoke to soon. It can't be simplified the way I thought. I will think about it some more.

 

[olgerm]

I think it's a mistake to introduce the components of the vectors into the calculation. Note by the way that you're not allowed to make assumptions about how many of them there are.

I know ,but for example if I get $(a_1*b_1+a_2*b_2+a_3*b_3)*(a_1*c_1+a_2*c_2+a_3*c_3)*(c_1*b_1+c_2*b_2+c_3*b_3)$ then I can just check is $(\vec a \cdot \vec b) (\vec a \cdot \vec c)(\vec b \cdot \vec c)$ equal to

$\mid \vec b -\frac{\vec a*\vec b}{\vec a*\vec c} * \vec c \mid$

or not.

 

[SammyS]

...

@olgerm, there is an interesting relationship between the two expressions (Q1 and Q2). Can you see how to get the second from the first?

(Edited by SammyS for readability.)
$\mid \vec b -B \vec c \mid$

1. $\displaystyle B=\frac{\vec a\cdot \vec b}{\vec a\cdot \vec c}$
2. $\displaystyle B=\frac{(\vec a\cdot \vec c) (\vec b\cdot \vec b)-(\vec a\cdot \vec b) (\vec b\cdot \vec c)}{(\vec a\cdot \vec b)(\vec c\cdot \vec c)-(\vec a\cdot \vec c(\vec b\cdot \vec c)}$

$\sqrt{(\vec b\cdot \vec b)-2B(\vec b\cdot \vec c)+B^2(\vec c\cdot \vec c)}$

Rather than that, substitute $\ \vec b - \vec c \$ in for $\ \vec b \$ in $\displaystyle \ \left |\, \vec b -\frac{\vec a\cdot \vec b}{\vec a\cdot \vec c} \, \vec c \, \right| \$.

It seems to me that the vector triple product gives a form which is at least visually simplified.

 

[Fredrik]

I don't see anything better than to use the vector triple product as discussed above. Look up that term at Wikipedia or something, find a formula that contains an expression that looks a bit like what you have, and then try to use that formula. I'm not sure why that result should be considered "simpler" though.