# algebraic definition of vector product

I am struggling to justify the the consistency of algebraic definition of vector products.

Say given Two vector A , B

A.B = |A|. |B| cos(0)


where Lets assume

A is represented by (x1,y1)
B is represented by (x2,y2)
|A| = sqrt(x1^2 + y1^2)
|B| = sqrt(x2^2 + y2^2)


so by Algebraic definition of dot product

A.B = x1.x2 + y1.y2


so

cos(0) = A.B / (sqrt(x1^2 + y1^2) * sqrt(x2^2 + y2^2))

which is equivalent to

  cos(0) =  (x1.x2 + y1.y2) / (sqrt(x1^2 + y1^2) *  sqrt(x2^2 + y2^2))


if two vector are parallel

cos(0) should be one 1


for two parallel vectors ratio same ratio (which essentially is the slope in 2D ) x1/y1 == x2/y2

So for two parallel vector how to you prove

(x1.x2 + y1.y2) / (sqrt(x1^2 + y1^2) +  sqrt(x2^2 + y2^2)) equals to 1

Given
x1/y1 == x2/y2

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Do you mean |B| = sqrt(x2^2 + y2^2)? $\;$ –  Ricky Demer Sep 9 '13 at 22:55
@Rick Yes. i am representing the vector B as (x2,y2) –  David Sep 9 '13 at 22:56
@Rick Fixed the typo. –  David Sep 9 '13 at 22:57
@David The $+$ sign in the denominator connecting two square roots should be multiplication. –  Tunococ Sep 10 '13 at 0:25

You turned the product of $|A|$ and $|B|$ into a sum of the square roots. Replacing with the original product, you get this:

There is a number $k$ such that $y_1=kx_1$ and $y_2=kx_2$, since $x=x_1/y_1=x_2/y_2$.

Now your big expression is equal to $\frac{x_1x_2+k^2x_1x_2}{\sqrt{x_1^2+k^2x_1^2}\sqrt{x_2^2+k^2x_2^2}}=\frac{(1+k^2)x_1x_2}{|x_1|\sqrt{1+k^2}|x_2|\sqrt{1+k^2}}=\frac{x_1x_2}{|x_1x_2|}$.

Notice that you get $1$ or $-1$, which is normal, because your assumption $x_1/y_1=x_2/y_2$ does not imply that the two vectors have the same direction: they can be opposite.

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\begin{align} \frac{x_1 x_2 + y_1 y_2}{\sqrt{x_1^2 + y_1^2}\sqrt{x_2^2 + y_2^2}} = \frac{\frac{x_1}{y_1}\frac{x_2}{y_2} + 1}{\sqrt{\left(\frac{x_1}{y_1}\right)^2 + 1}\sqrt{\left(\frac{x_2}{y_2}\right)^2 + 1}} = \frac{\left(\frac{x_1}{y_1}\right)^2 + 1}{\sqrt{\left(\frac{x_1}{y_1}\right)^2 + 1}\sqrt{\left(\frac{x_1}{y_1}\right)^2 + 1}} = 1 \end{align}

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If two non-zero vectors $A,B$ are parallel, then there is some constant $\lambda$ such that $A=\lambda B$.

Then $\cos \theta = \frac{A \cdot B}{\|A\| \|B\|} = \frac{\lambda B \cdot B}{|\lambda|\|B\| \|B\|} = \frac{\lambda}{|\lambda|} \frac{\|B\|^2}{\|B\|^2} = \operatorname{sgn} \lambda$.

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