# Thread: Proof

1. ## Proof

Hi,
This is my first post on the forum, but I have been reading this forum for at least 2 years now. It has always been very useful to read some of the posts you guys make.

However, now, it seems like I can't solve my problem simply by reading the forum...I would need a little extra help.

Here is how it goes:

(Note: Let bold letters be vectors)
I know that u=(u1,u2,u3), v=(v1,v2,v3) and w=(w1,w2,w3)
Let A=$\displaystyle \begin{array}{ccc}u1&u2&u3\\v1&v2&v3\\w1&w2&w3\end {array}$

I have to prove that:
(Note: Let u,v and w, be vectors)
det(A) $\displaystyle \leq$ $\displaystyle \parallel u \parallel \parallel v \parallel \parallel w \parallel$

I tried to replace the norm of u times norm of v times norm of w by its equivalent using their values (u1,u2,u3), (v1,v2,v3), (w1,w2,w3).

Hence, $\displaystyle \parallel u \parallel = \sqrt{u \cdot u}$ ...

But then, all I got was a huge equation and the strange impression that I am not even close from doing the right thing...

Anyone could help me head in the right direction ?

Thanks a lot !

2. Originally Posted by elitethut
Hi,
This is my first post on the forum, but I have been reading this forum for at least 2 years now. It has always been very useful to read some of the posts you guys make.

However, now, it seems like I can't solve my problem simply by reading the forum...I would need a little extra help.

Here is how it goes:

(Note: Let bold letters be vectors)
I know that u=(u1,u2,u3), v=(v1,v2,v3) and w=(w1,w2,w3)
Let A=$\displaystyle \begin{array}{ccc}u1&u2&u3\\v1&v2&v3\\w1&w2&w3\end {array}$

I have to prove that:
(Note: Let u,v and w, be vectors)
det(A) $\displaystyle \leq$ $\displaystyle \parallel u \parallel \parallel v \parallel \parallel w \parallel$

I tried to replace the norm of u times norm of v times norm of w by its equivalent using their values (u1,u2,u3), (v1,v2,v3), (w1,w2,w3).

Hence, $\displaystyle \parallel u \parallel = \sqrt{u \cdot u}$ ...

But then, all I got was a huge equation and the strange impression that I am not even close from doing the right thing...

Anyone could help me head in the right direction ?

Thanks a lot !
The scalar triple product should do the trick.

$\displaystyle \vec{a} \cdot (\vec{b} \times \vec{c})=\begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$

Now just use the geometric defintions of the dot and cross product
$\displaystyle \vec{b} \times \vec{c}=||b|||c||\sin(\theta) \vec{n}$

where $\displaystyle \vec{n}$ is a unit vector normal to the plane formed by b and c $\displaystyle ||n||=1$

and $\displaystyle \vec{a} \cdot(\vec{b} \times \vec{c})=||a|||b||||c|\sin(\theta_1)\cos(\theta_2)$

and remember the sine and cosine are always less than or equal to 1.

3. Hmmm, I'm not sure I understand the role that $\displaystyle \vec{n}$ has to play in this.

4. Originally Posted by elitethut
Hmmm, I'm not sure I understand the role that $\displaystyle \vec{n}$ has to play in this.
When you cross two vectors you get a vector that is normal(perpendicular) to the plane spanned by the two vectors.

The n is just giving you the direction that the resulting vector is going.
The $\displaystyle ||b|||c||\sin(\theta)$is the length(magnitude) of the resulting vector.

This is needed becuase the dot product is a binary operation on two vectors, not a vector and magnitude. If we didn't have the unit normal the dot product would not make sense.

I hope this helps.

I have editied the above post for clarity as well.

5. Ok... So

$\displaystyle \vec{b} \times \vec{c}=||b|||c||\sin(\theta) \vec{n}$
and
$\displaystyle \vec{a}=||a||\cos(\theta) \vec{n}$

Therefore,
$\displaystyle \vec{a} \cdot(\vec{b} \times \vec{c})=||a|||b||||c|\sin(\theta_1)\cos(\theta_2)$

Is that right?

6. Originally Posted by elitethut
Ok... So

$\displaystyle \vec{b} \times \vec{c}=||b|||c||\sin(\theta) \vec{n}$
and
$\displaystyle \vec{a}=||a||\cos(\theta) \vec{n}$

Therefore,
$\displaystyle \vec{a} \cdot(\vec{b} \times \vec{c})=||a|||b||||c|\sin(\theta_1)\cos(\theta_2)$

Is that right?
$\displaystyle \vec{a}=||a||\cos(\theta) \vec{n}$

Is incorrect. a is just the vector a.
Now:
$\displaystyle d\cdot e=||d||||e||cos(\theta)$ where $\displaystyle \theta$ is the angle beteween d and e.

Therefore:
$\displaystyle a\cdot(b\times c)=||a||\Bigl(||\bigl(||b|||c||\sin(\theta) \vec{n} \bigr)||\Bigr)\cos(\theta_2)=||a||||b||||c||\sin(\ theta)\cos(\theta_2)||n||$

Now n is a unit vector so:

$\displaystyle ||a||||b||||c||\sin(\theta)\cos(\theta_2)||n||=||a ||||b||||c||\sin(\theta)\cos(\theta_2)$

Be careful, don't confuse $\displaystyle \theta$ and $\displaystyle \theta_2$ together.
The one inside sin is the angle between b and c, and the one inside cos is the angle between a and the resulting vector of (bxc).

7. Ok,

So basically,

$\displaystyle ||a||||b||||c||\sin(\theta)\cos(\theta_2)$
is equal to Det(A)...but how does this prove that the det(A) is smaller than $\displaystyle ||a|||b|||c||$

Edit: Oh, is it simply that $\displaystyle \sin(\theta)$ and $\displaystyle \cos(\theta_2)$ will always give values bellow 1 (or equal to one), and therefore the value of $\displaystyle ||a||||b||||c||$ times 2 values that are equal or smaller to 1 will always give a smaller or equal value than $\displaystyle ||a||||b||||c||$

8. Originally Posted by elitethut
Edit: Oh, is it simply that $\displaystyle \sin(\theta)$ and $\displaystyle \cos(\theta_2)$ will always give values bellow 1 (or equal to one), and therefore the value of $\displaystyle ||a||||b||||c||$ times 2 values that are equal or smaller to 1 will always give a smaller or equal value than $\displaystyle ||a||||b||||c||$
Yes, that is correct.

9. Ok, I understand everything except one thing.

Isn't the formula supposed to be
$\displaystyle ||\vec{b} \times \vec{c}||=||b|||c||\sin(\theta) \vec{n}$

and not :

$\displaystyle \vec{b} \times \vec{c}=||b|||c||\sin(\theta) \vec{n}$ ???

10. Originally Posted by elitethut
Ok, I understand everything except one thing.

Isn't the formula supposed to be
$\displaystyle ||\vec{b} \times \vec{c}||=||b|||c||\sin(\theta) \vec{n}$

and not :

$\displaystyle \vec{b} \times \vec{c}=||b|||c||\sin(\theta) \vec{n}$ ???
$\displaystyle \vec{b} \times \vec{c}$ is a vector while $\displaystyle ||\vec{b} \times \vec{c}||$ is a length of the vector.

The length of the vector that is the result of a cross product $\displaystyle \vec{b} \times \vec c$ is
$\displaystyle ||\vec{b} \times \vec c||=||b|||c||\sin(\theta)$,
but the vector it self is $\displaystyle ||b|||c||\sin(\theta) \vec{n}$

Read again what TheEmptySet posted above:
When you cross two vectors you get a vector that is normal(perpendicular) to the plane spanned by the two vectors.

The n is just giving you the direction that the resulting vector is going.
The $\displaystyle ||b|||c||\sin(\theta)$is the length(magnitude) of the resulting vector.

This is needed becuase the dot product is a binary operation on two vectors, not a vector and magnitude. If we didn't have the unit normal the dot product would not make sense.

I hope this helps.

I have editied the above post for clarity as well.