# dot and cross products

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• Jan 7th 2009, 10:02 AM
Skerven
dot and cross products
I just want to know how the geometric, algebraic, and trigonometric definitions of the dot and cross products of two vectors relate to each other.

The dot product is defined as both the sum of squares of components of two vectors and also the product of their lengths times cos(θ). I also know that the cross product is the determinant of the matrix with unit vectors i,j, and k in the first row, the components of a in the second, and that of b in the third and also the product of their lengths times sin(θ) times a unit vector n that is perpendicular to both a & b.

But how does:

$\displaystyle a_1*b_1 + a_2*b_2 + a_3*b_3=\sqrt{a_1^2 + a_2^2 + a_3^2}*\sqrt{b_1^2 + b_2^2 + b_3^2}*\cos{\theta}$

and the determinant of $\displaystyle \begin{bmatrix}i&j&k\\a_1&a_2&a_3\\b_1&b_2&b_3\end {bmatrix}=\sqrt{a_1^2 + a_2^2 + a_3^2}*\sqrt{b_1^2 + b_2^2 + b_3^2}*\sin{\theta}*n$

I already understand the geometric definition via example (Work = Force*Distance and Torque), and you don't have to explain it to me algebraically; as long as the fundamental idea is explained, any way is fine, thanks.
• Jan 7th 2009, 10:39 AM
Mush
Quote:

Originally Posted by Skerven
I just want to know how the geometric, algebraic, and trigonometric definitions of the dot and cross products of two vectors relate to each other.

The dot product is defined as both the sum of squares of components of two vectors and also the product of their lengths times cos(θ). I also know that the cross product is the determinant of the matrix with unit vectors i,j, and k in the first row, the components of a in the second, and that of b in the third and also the product of their lengths times sin(θ) times a unit vector n that is perpendicular to both a & b.

But how does:

$\displaystyle a_1*b_1 + a_2*b_2 + a_3*b_3=\sqrt{a_1^2 + a_2^2 + a_3^2}*\sqrt{b_1^2 + b_2^2 + b_3^2}*\cos{\theta}$

and the determinant of $\displaystyle \begin{bmatrix}i&j&k\\a_1&a_2&a_3\\b_1&b_2&b_3\end {bmatrix}=\sqrt{a_1^2 + a_2^2 + a_3^2}*\sqrt{b_1^2 + b_2^2 + b_3^2}*\sin{\theta}*n$

I already understand the geometric definition via example (Work = Force*Distance and Torque), and you don't have to explain it to me algebraically; as long as the fundamental idea is explained, any way is fine, thanks.

Two vectors have different angles and magnitudes. When you combine these vectors you expect the combination to be a sort of 'average' of the the angle between the two original vectors, and you expect the magnitude to increase/decrease according to the magnitudes of the originals.

When you multiply two vectors using the cross product you are doing the same thing. The only real difference being the inclusion of the unit vector n, which merely specifies that the cross product of two vectors is normal to both.
• Jan 7th 2009, 04:24 PM
Skerven
For the dot product I'm stuck. Where c is the length of the projection of a onto b...
$\displaystyle a_1*b_1 + a_2*b_2 = \sqrt{a_1^2 + a_2^2}*\sqrt{b_1^2 + b_2^2}*\cos{\theta}$
$\displaystyle a_1*b_1 + a_2*b_2 = \frac{\sqrt{a_1^2 + a_2^2}*\sqrt{b_1^2 + b_2^2}*c}{\sqrt{a_1^2 + a_2^2}}$
$\displaystyle a_1*b_1 + a_2*b_2 = \sqrt{b_1^2 + b_2^2}*c$
• Jan 7th 2009, 04:26 PM
Mush
Quote:

Originally Posted by Skerven
And of the dot product?

The first paragraph of my post was referring to the dot product.
• Jan 7th 2009, 05:05 PM
Skerven
Quote:

Originally Posted by Mush
The first paragraph of my post was referring to the dot product.

Oh, I thought you were talking about vector addition... I've already figured out the cross product. The determinant of the so-called matrix with i,j, and k in the first row is just a matter of splitting up the two vectors in terms of the standard basis (i, j, and k), and then using the properties of the cross product (which I understand) to simplify. That matrix is just a representation.

But I still don't understand how the mathematicians who first came up with the definition of the dot product figured out that

$\displaystyle a_1*b_1 + a_2*b_2 + a_3*b_3=\sqrt{a_1^2 + a_2^2 + a_3^2}*\sqrt{b_1^2 + b_2^2 + b_3^2}*\cos{\theta}$
• Jan 8th 2009, 12:11 AM
Opalg
Quote:

Originally Posted by Skerven
I still don't understand how the mathematicians who first came up with the definition of the dot product figured out that

$\displaystyle a_1*b_1 + a_2*b_2 + a_3*b_3=\sqrt{a_1^2 + a_2^2 + a_3^2}*\sqrt{b_1^2 + b_2^2 + b_3^2}*\cos{\theta}$

It's not so mysterious if you look at it in two dimensions. Suppose that the vectors are $\displaystyle \mathbf{x} = (x_1,x_2)$ and $\displaystyle \mathbf{y} = (y_1,y_2)$. Using polar coordinates, we can write $\displaystyle x_1 = r\cos\theta$, $\displaystyle x_2 = r\sin\theta$, $\displaystyle y_1 = s\cos\phi$, $\displaystyle y_2 = s\sin\phi$, where $\displaystyle r = \sqrt{x_1^2+x_2^2}$, $\displaystyle s = \sqrt{y_1^2+y_2^2}$, and $\displaystyle \theta,\ \phi$ are the angles that the vectors make with the x-axis.

Then $\displaystyle \mathbf{x.y} = x_1y_1+x_2y_2 = rs(\cos\theta\cos\phi+\sin\theta\sin\phi) = rs\cos(\theta-\phi)$. So the dot product (as defined algebraically) is equal to the product of the lengths of the vectors times the cosine of the angle between them.

That argument doesn't work so neatly in three dimensions. But the subspace spanned by two vectors is two-dimensional, and with a bit of work you can show that the (algebraically defined) dot product in three-dimensional space coincides with the two-dimensional dot product on any two-dimensional subspace.
• Jan 8th 2009, 08:07 AM
Skerven
Quote:

Originally Posted by Opalg
Then $\displaystyle \mathbf{x.y} = x_1y_1+x_2y_2 = rs(\cos\theta\cos\phi+\sin\theta\sin\phi) = rs\cos(\theta-\phi)$. So the dot product (as defined algebraically) is equal to the product of the lengths of the vectors times the cosine of the angle between them.

The first place I should have looked was my Linear Algebra book. Mr. Gilbert Strang mentioned that the definition with the cosine in it (too lazy to write in Latex now) could be solved with the Law of Cosines. Then I looked at my Calculus 3 book and there you have it, the proof with the Law of Cosines. Yours is another proof that uses a similar trig identity.

Quote:

Originally Posted by Opalg
That argument doesn't work so neatly in three dimensions. But the subspace spanned by two vectors is two-dimensional, and with a bit of work you can show that the (algebraically defined) dot product in three-dimensional space coincides with the two-dimensional dot product on any two-dimensional subspace.

I just found out that the sum of the squares of the cosines of the angles created by a 3-dimensional vector with the x-,y- and z-axes equals 1.
$\displaystyle \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$