How can I use trigonometric functions to simplify my vector algebra problem?

[The_Brain]

I hate Vector Algebra!

I am having some trouble doing some vector algebra and any help or direction is greatly appreciated.

In our lecture we drew a vector C(s) = A + s(B-A) which described a line passing through points A and B and found that the vector of min. length (shortest dist. from origin to line) was given by s_o = [A (dot) (A-B)]/[|B-A|². The actual problem is to plug s_o into C(s) to show that C(s_o) = (sqrt(|A|²|B|² - (A dot B)²))/(|B-A|).

What I have so far consists of changing [A (dot) (A-B)]/[|B-A|² to [-A (dot) (B-A)]/[|B-A|² so I can square the (B-A) however, when I do that I do not know how to then multiply that by the dot product of A. I know that A dot (B-A) is A dot B minus A dot A (A squared) but I have no idea how to do A dot (B-A)². Thanks for any help.

 

[robphy]

... I have no idea how to do A dot (B-A)². Thanks for any help.

For a vector V, the quantity V²= ( V (dot) V ) is a scalar.
Note that A (dot) ( (B-A)² ) is not a legal operation.

 

[The_Brain]

Sorry about my notation - I should have been more obvious about the A squared part being a scalar but thanks for telling me that A (dot) ( (B-A)2 ) is not a legal operation, now I know not to work in that direction.

 

[robphy]

Maybe not the most elegant solution

$$\def\vA {\vec A} \def\vB {\vec B} \def\vC {\vec C} \begin{align*} \vC &= \vA + s(\vB-\vA)\\ C^2 &= A^2 + 2s\vA\cdot(\vB-\vA)+s^2(\vB-\vA)\cdot(\vB-\vA)\\ \end{align*}$$

with
$$\def\vA {\vec A} \def\vB {\vec B} \def\vC {\vec C} s_0=\frac{\vA\cdot(\vA-\vB)}{|B-A|^2}$$

$$\def\vA {\vec A} \def\vB {\vec B} \def\vC {\vec C} \def\vAB {(\vA\cdot\vB)} \begin{align*} C(s_0)^2 &= A^2 + 2 \left[ \frac{\vA\cdot(\vA-\vB)}{|B-A|^2}\right]\vA\cdot(\vB-\vA)+\left[ \frac{\vA\cdot(\vA-\vB)}{|B-A|^2}\right]^2(\vB-\vA)\cdot(\vB-\vA)\\ &= A^2 - 2 \left[ \frac{\vA\cdot(\vA-\vB)}{|B-A|^2}\right]\vA\cdot(\vA-\vB)+ \frac{(\vA\cdot(\vA-\vB))^2}{|B-A|^2}\\ &= A^2 - \frac{(\vA\cdot(\vA-\vB))^2}{|B-A|^2}\\ &= \frac{A^2 (\vB-\vA)\cdot(\vB-\vA)- (\vA\cdot(\vA-\vB))^2}{|B-A|^2}\\ &= \frac{A^2 (B^2-2\vAB+A^2)- (A^2-\vAB)^2}{|B-A|^2}\\ &= \frac{A^2 (B^2-2\vAB+A^2)- (A^4-2A^2\vAB +\vAB^2)}{|B-A|^2}\\ &= \frac{A^2 B^2-2A^2\vAB+A^4- A^4+2A^2\vAB -\vAB^2}{|B-A|^2}\\ &= \frac{A^2 B^2-\vAB^2}{|B-A|^2}\\ \end{align*}$$

 

[The_Brain]

I think that's as elegenat as it can get. After about an hour two days ago, I finally came up with a solution that was just as long if not longer than yours. It seems as if it's just brute force is the route here. Thanks for all your help.

 

[robphy]

If you allow yourself to use some trigonometric functions, it may have been easier to manipulate.
Note that $\vec A\cdot \vec B=AB\cos\theta$ so that $A^2B^2-(\vec A\cdot \vec B)^2=A^2B^2-A^2B^2\cos^2\theta=A^2B^2(1-\cos^2\theta)=A^2B^2\sin^2\theta$.