Proof

**elitethut** · October 26th 2009, 05:32 PM

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= $\begin{array}{ccc}u1&u2&u3\\v1&v2&v3\\w1&w2&w3\end {array}$

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

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

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 !

**TheEmptySet** · October 26th 2009, 06:33 PM

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= $\begin{array}{ccc}u1&u2&u3\\v1&v2&v3\\w1&w2&w3\end {array}$

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

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

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.

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

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

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

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

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

**elitethut** · October 27th 2009, 01:22 PM

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

**TheEmptySet** · October 27th 2009, 01:40 PM

Originally Posted by elitethut

Hmmm, I'm not sure I understand the role that $\vec{n}$ 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 $||b|||c||\sin(\theta)$ 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.

**elitethut** · October 27th 2009, 03:42 PM

Ok... So

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

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

Is that right?

**hjortur** · October 27th 2009, 05:06 PM

Originally Posted by elitethut

Ok... So

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

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

Is that right?

$ \vec{a}=||a||\cos(\theta) \vec{n} $

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

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

Now n is a unit vector so:

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

Be careful, don't confuse $\theta$ and $\theta_2$ 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).

**elitethut** · October 27th 2009, 05:32 PM

Ok,

So basically,

$||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 $ ||a|||b|||c|| $

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

**hjortur** · October 28th 2009, 03:56 AM

Originally Posted by elitethut

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

Yes, that is correct.

**elitethut** · October 28th 2009, 12:32 PM

Ok, I understand everything except one thing.

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

and not :

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

**hjortur** · October 28th 2009, 01:28 PM

Originally Posted by elitethut

Ok, I understand everything except one thing.

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

and not :

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

$\vec{b} \times \vec{c}$ is a vector while $||\vec{b} \times \vec{c}||$ is a length of the vector.

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

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 $||b|||c||\sin(\theta)$ 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.

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