The coordinates of points $P_1, P_2 $ and angle $\theta$ are known and the sides $ PP_1$ and $PP_2$ have equal length. Although the point $P$ can be calculated using isosceles triangle properties and Pythagoras Theorem, I like to try vector method.
Deriving from the equation,
$$ cos(\theta) = n_1 \cdot n_2 ,$$
where
$ n_1 = \frac{\vec{PP_1}}{|\vec{PP_1}|}$
$ , n_2 = \frac{\vec{PP_2}}{|\vec{PP_2}|}$
$$ cos(\theta) = \frac{P_1 – P}{\sqrt{(P_{1x} – P_x)^2 + (P_{1y} – P_y)^2}}
\cdot \frac{P_2-P}{\sqrt{(P_{2x}-P_x)^2 + (P_{2y} – P_y)^2}}$$
How can I find $P$ when $P_1$, $P_2$ and $cos(\theta)$ values are substituted in the above equation? I get confused whether it is correct to solve the equation by substituting the vectors with coordinates if the dot product is only for vectors.
Best Answer
Let $P_1(x_1,y_1)$, $P_2(x_2,y_2)$ and $D$ is a midpoint of $P_1P_2$.
Thus, $$\vec{P_1P}=\vec{P_1D}+\vec{DP}=\frac{1}{2}(x_2-x_1,y_2-y_1)\pm\frac{1}{2}\cot\frac{\theta}{2}(y_2-y_1,x_1-x_2),$$ which says $$P\left(\frac{1}{2}\left(x_2-x_1\pm\cot\frac{\theta}{2}(y_2-y_1\right)+x_1,\frac{1}{2}\left(y_2-y_1\pm\cot\frac{\theta}{2}(x_1-x_2\right)+y_1\right)$$ or $$P\left(\frac{1}{2}\left(x_2+x_1\pm\cot\frac{\theta}{2}(y_2-y_1\right),\frac{1}{2}\left(y_2+y_1\pm\cot\frac{\theta}{2}(x_1-x_2\right)\right)$$