1. ## [SOLVED] Complicated Vector Problem.

The Problem is:

"A horseman left the village A at point $(1,4)$ and began riding along a straight road whose direction was given by the vector v=7i+4j. The at some point he turned at a right angle; He never changed direction again until he arrived in the village [LaTeX ERROR: Convert failed] at the point with the coordinates (8,13). At the point where he made the turn he buried a jar full of silver coins, Unfortunately, he forgot the coordinates of the point. Find the point.

So I drew the graph and labeled the point where he changed direction as $C$ with the coordinates $(x,y)$. This means the vector $CA$ is <x-1, y-4> and the vector $BC$ is <8-x,13-y>.

Next, I worked out the dot product of the two vectors to be (x-1)(8-x)+(y-4)(13-y)=0 It has to be equal to 0 since the angle is a right angle and it is orthogonal. Been working on this problem for about 45 minutes, and have no idea how to solve it. Any help is appreciated.

2. ## Re: Complicated Vector Problem.

Think in a triangle. Try doing the problem in terms of projections. You can tell that the vector $AX$ ( $X=(x,y)$ the unknown point) is the projection of $AB$ onto the direction vector $v=\{7,4\}$ (since there is a right angle).
So you write: $AX=proj_{\{7,4\}}AB$. Then you have $AX=\left(\frac{\{7,9\}\cdot \{7,4\}}{\{7,4\}\cdot \{7,4\}}\right)\{7,4\}$.
Solving this results in $AX = \dfrac{17}{13}\cdot\{7,4\}=\left\{\dfrac{119}{13}, \dfrac{68}{13} \right\}$.
As $AX=X-A\Rightarrow X=\left\{\dfrac{119}{13}, \dfrac{68}{13} \right\}+\{1,4\}=\left\{\dfrac{132}{13}, \dfrac{120}{13} \right\}$.

Now you could use the fact that $AX\cdot XB=0$ to check if the answer is correct.

3. ## Re: Complicated Vector Problem.

Thanks! I didn't think about it in terms of projections. This helps a lot.