Finding coordinates of a third point using vector algebra Sorry I didn't know how to make the title better to represent the question properly, but here it goes. You have 3 trees on an island. Let's call them A,B and D. You walk from D to A, then you turn 90∘ to the left and walk the same distance, and then mark that point, let's call it K1. Then, you walk from D to B, then you turn 90∘ to the right and walk the same distance and mark that point as K2. In the middle point between K1 and K2 is the treasure! However, when you go to the island, you see that D tree is gone! How can you find the treasure by using vector algebra? (using complex numbers is not allowed)
 A: Let’s do this first using the prohibited complex numbers, as the solution is a bit simpler when expressed that way. We can then translate it into vectors.  
Multiplying $i$ rotates a complex number 90° counterclockwise, and multiplying by $-i$ rotates it clockwise. Using this we have $$K_1=A+(A-D)\,i \\ K_2=B-(B-D)\,i$$ and their midpoint is $$\begin{align}X=\frac12(K_1+K_2)&=\frac12[A+(A-D)\,i+B-(B-D)\,i] \\ &=\frac12(A+B)+\frac12(A-B)\,i.\end{align}$$ The location of the treasure is independent of $D$, so we can find the treasure without this tree, as suggested by the problem. The resulting equation for $X$ tells us where to dig: go halfway from $B$ to $A$, turn left, and go the same distance.  
The above calculations translate directly into vector terms. Let $R$ represent a 90° counterclockwise rotation. Then we have $$K_1=A+R(A-D)=(I+R)A-RD \\ K_2=B+R^{-1}(B-D)=(I+R^{-1})B-R^{-1}D$$ and $$X=\frac12[(I+R)A-RD+(I+R^{-1})B-R^{-1}D].$$ Now, since these are 90° rotations, then for any vector $V$, $R^{-1}V=-RV$, so this equation becomes $$\begin{align}X &= \frac12[(I+R)A-RD+(I-R)B+RD] \\ &=\frac12[A+RA+B-RB] \\ &= \frac12(A+B)+\frac12R(A-B).\end{align}$$ Finally, $R(x,y)=(-y,x)$, so in terms of coordinates, $$X=\frac12(x_A+x_B,y_A+y_B)+\frac12(y_B-y_A,x_A-x_B).$$
