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The cross product $\vec{a} \times \vec{b}$ can be written as the determinant of the matrix:

$$\left| \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ a_i & a_j & a_k \\ b_i & b_j & b_k \end{matrix}\right|$$

Is there any physical significance to this matrix, or is it just some mathematical trick?

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Agha, your question has already attracted one Move to MathSE vote, so I've tried to rewrite it in a way more applicable to the PhysicsSE. However this is a big change. If you don't like my change please roll it back. –  John Rennie Aug 31 at 5:44
    
@JohnRennie thanks a lot for that edit john –  agha rehan abbas Aug 31 at 7:36
    
There's a very nice post on math.SE: What's an intuitive way to think about the determinant? Take a look at how the properties of the determinant uniquely determine its mathematical form. I suspect this argument can be related to the cross product, but it's late here. –  BMS Aug 31 at 9:56
    
I always look at the cross product as a linear operator $$ \vec{a}\times\vec{b} = \begin{pmatrix}0&-a_k&a_j\\a_k&0&-a_i\\-a_j&a_i&0\end{pmatrix}\begin{pmatrix}b_i\\b‌​_j\\b_k\end{pmatrix}$$ –  ja72 Aug 31 at 18:00
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What I mean is that the $[\vec{a}\times]$ matrix transforms the $\vec{b}$ vector. For example in a rotating frame $[\vec{\omega}\times] \vec{r}$ returns the rate of change of the components of a body fixed vector $\vec{r}$. The $\times$ takes a vector and turns it into an operator (like a function but in vector space). –  ja72 Aug 31 at 22:32

3 Answers 3

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Is there any physical significance to this matrix

The physical (geometric) relevance to the matrix

$$\left| \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ a_i & a_j & a_k \\ b_i & b_j & b_k \end{matrix}\right|$$

with regard to the cross product $\vec{a} \times \vec{b}$ is

1: that the three vectors $\vec{i}$, $\vec{j}$, and $\vec{k}$ constitute a vector basis that spans a space which is

  • either also spanned by $\vec{a}$, $\vec{b}$, and one additional vector which is perpendicular to $\vec{a}$ as well as $\vec{b}$;

  • or, in case that vectors $\vec{a}$ and $\vec{b}$ are parallel to each other, also spanned by $\vec{a}$ and two additional (non-parallel) vectors.

2: the three basis vectors $\vec{i}$, $\vec{j}$, and $\vec{k}$ are pairwise orthogonal (perpendicular) to each other.

Therefore vectors $\vec{a}$ and $\vec{b}$ as well as the cross product vector $\vec{a} \times \vec{b}$ can be completely and uniquely expressed in terms of the corresponding components:

$\vec{a} := a_i \vec{i} + a_j \vec{j} + a_k \vec{k}$,
$\vec{b} := b_i \vec{i} + b_j \vec{j} + b_k \vec{k}$, and

$\vec{a} \times \vec{b} := \{ab\}_i \vec{i} + \{ab\}_j \vec{j} + \{ab\}_k \vec{k}$.

Finally: 3: the three basis vectors $\vec{i}$, $\vec{j}$, and $\vec{k}$ have equal magnitudes:

$| \vec{i} | = | \vec{j} | = | \vec{k} |$.

As a consequence, the "mathematical trick" of expressing the cross product vector $\vec{a} \times \vec{b}$ as the above determinant "works":

The component of cross product vector $\vec{a} \times \vec{b}$ "along/parallel to" vector $\vec{a}$ vanishes explicitly:

$$\left( a_i (a_j b_k - a_k b_j) \frac{(| \vec{i} |)^2}{| \vec{a} \times \vec{b} |} \right) + \left( a_j (a_k b_i - a_i b_k) \frac{(| \vec{j} |)^2}{| \vec{a} \times \vec{b} |} \right) + \left( a_k (a_i b_j - a_j b_i) \frac{(| \vec{k} |)^2}{| \vec{a} \times \vec{b} |} \right) = $$

$$\frac{(| \vec{i} |)^2}{| \vec{a} \times \vec{b} |} \left( a_i (a_j b_k - a_k b_j) + a_j (a_k b_i - a_i b_k) + a_k (a_i b_j - a_j b_i) \right) = 0,$$

and likewise the component of cross product vector $\vec{a} \times \vec{b}$ "along/parallel to" vector $\vec{b}$ vanishes explicitly;
i.e. cross product vector $\vec{a} \times \vec{b}$ is expressed explicitly orthogonal to both vectors $\vec{a}$ and $\vec{b}$.

And, no less important the magnitude of cross product vector $\vec{a} \times \vec{b}$ "comes out correctly", i.e. such that

$$ \Big( | \vec{a} \times \vec{b} | \Big)^2 := $$ $$ \left( (a_j b_k - a_k b_j)^2 + (a_k b_i - a_i b_k)^2 + (a_i b_j - a_j b_i)^2 \right) ~ (| \vec{i} |)^4 = $$ $$ \left( (a_i^2 + a_j^2 + a_k^2) ~ (b_i^2 + b_j^2 + b_k^2) \right) ~ (| \vec{i} |)^4 - \left( (a_i b_i + a_j b_j + a_k b_k) ~ (| \vec{i} |)^2 \right)^2 := $$ $$ \Big( | \vec{a} | \Big)^2 \Big( | \vec{b} | \Big)^2 - | \vec{a} | ~ | \vec{b} | ~ a_b ~ b_a,$$

where $b_a$ denotes the component of vector $\vec{b}$ "along/parallel to" vector $\vec{a}$, and $a_b$ denotes the component of vector $\vec{a}$ "along/parallel to" vector $\vec{b}$.

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thanks your answer helped me specially –  agha rehan abbas Aug 31 at 13:24
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The last term in the last formula, $(a_bb_a)^2$, seems to be a mistake. By your definition $a_b=a\cos\alpha$ and $b_a=b\cos\alpha$ with $\alpha$ being the angle between vectors, so $a_bb_a=ab\cos^2\alpha$, the power of cosine being 1 more than needed. –  firtree Aug 31 at 14:30
    
firtree: "The last term in the last formula, [...] seems to be a mistake." -- That's right, thanks! Explicitly: $$ \Big(|\vec{a}\times\vec{b}|\Big)^2:=$$ $$ \left((a_j b_k-a_k b_j)^2+(a_k b_i-a_i b_k)^2+(a_i b_j-a_j b_i)^2\right)~(|\vec{i}|)^4=$$ $$\left((a_i^2+a_j^2+a_k^2)~(b_i^2+b_j^2+b_k^2)\right)~(|\vec{i}|)^4-\left((a_i b_i+a_j b_j+a_k b_k)~(|\vec{i}|)^2\right)^2:=$$ $$\Big(|\vec{a}|\Big)^2~\Big(|\vec{b}|\Big)^2-|\vec{a}|~| \vec{b}|~a_b~b_a;$$ where I've tried hard, of course, not to introduce any notions (such as "angle") which don't appear in the question itself. –  user12262 Aug 31 at 21:10

The cross product is defined to be the vector which is perpendicular to both vectors, so for instance the force exerted on a rod moving in a magnetic field is perpendicular to both its velocity and the field, hence is given by their cross product.

Now if you work out which vector is perpendicular to both vectors you get the determinant of the two vectors (In other words, writing it as a determinant is only to make it more easy on the eye, I don't think there is a deep reason to it)

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To make it easy to see the logic behind, imagine that the first vector is directed to $\vec{i}$ and the second to $\vec{j}$. You get

$$\left| \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ a_i & 0 & 0 \\ 0 & b_j & 0 \end{matrix}\right|$$

and according to Wikipedia the product in this case is $ab\vec{k}$.

There are some physical phenomena which can be described with vector products such as electromagnetic induction. More generally is the example where two forces in space act on a body and no one is directed to the axis of your coordinate system. Of course you are free to rotate your coordinate system to one of the forces and you get an improved calculation, that has more zeros in the determinant.

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