Problem with vectors, points and intersection

Question

I have a problem for an online assignment which I was able to complete. The problem was:

Gandalf the Grey started in the Forest of Mirkwood at a point with coordinates (−1,3) and arrived in the Iron Hills at the point with coordinates (1,6). If he began walking in the direction of the vector v=5i+1j and changes direction only once, when he turns at a right angle, what are the coordinates of the point where he makes the turn.

I was able to solve the problem by moving the the initial coordinates to the origin and hence moving the other at (2,3). Then, I took the inverse of the speed (which is the normal as well) and got w=1i-5j.

From there, I went freestyle and found the y=x+b version of the vectors, solved for x and y and moved them back to the original place finishing with the turn being at (1.5,3.5).

I got the right answer so I guess it works but would you guys have made it differently? I'm quite new to that kind of stuff so I couldn't think of a way of doing it with vectors and points only.

Thanks!

Answer 1

I think translation of the axes is totally unnecessary. In particular, his coordinates before turning can be described as $x = -1 + 5a$ and $y=3+a$ (which implies $5y=x+16$ if you prefer to not work with parametric equations).

Now, we need to compute the coordinates of the point where he turns at the right angle. Obviously, the vector from that point to $(1,6)$ is perpendicular to $5 \vec i + 1 \vec j$. Consequently, $((2-5a)\, \vec i + (3-a)\, \vec j)\cdot(5\vec i + \vec j)=0 \implies a=1/2 \implies (x,y)=(3/2,7/2)$.

Answer 2

A simple way to solve this is to use the vector dot product as in Math Lover’s answer, but you can also solve this using the Pythagorean theorem.

Let $p_0=(-1,3)$ and $p_1=(1,6)$ be the start and end points of the journey, respectively. The turning point $q=p_0+\alpha v$ for some $\alpha>0$. The two legs of the journey and the direct path between the end points form a right triangle, so we must have $$\|p_1-q\|^2+\|q-p_0\|^2=\|p_1-p_0\|^2$$ which when expanded and simplified becomes $$52\alpha^2-26\alpha=0.$$ The solutions to this equation are $\alpha = 0$ and $\alpha = \frac12$, so $q=\left(\frac32,\frac72\right)$.