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Bumbble Commhttps://math.techqa.club/user/bumbble-comm/detail
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As above diagrams shows I have two vector U and V. I want to determine weather V vector is in A side or the B side it means form the left side of the right side to the U vector.
I need some help to sort out this.
thanks in advance!
so... do you mean you want to know weather the point $\left(x_{2},y_{2}\right)$ is on the left or on the right of the line passing $\left(x_{1},y_{1}\right)$ and $\left(x_{3},y_{3}\right)$? In this case:
define a normal pointing always to the same side of your line passing the two points $\left(x_{1},y_{1}\right)$ and $\left(x_{3},y_{3}\right)$, that would mean any vector perpendicular on $\left(r_{x},r_{y}\right):=\left(x_{3},y_{3}\right)-\left(x_{1},y_{1}\right)$.
For instance $\left(s_{x},s_{y}\right):=\left(r_{y},-r_{x}\right)$ does it. this vector should point to the left if i'm not wrong :-) you better check this, i might mix up left and right.
then check the sign of the scalar product $\left(s_{x},s_{y}\right)\cdot\left(x_{2}-x_{1},y_{2}-y_{1}\right)$. if it's bigger than $0$ $\left(x_{2},y_{2}\right)$ is on the left, otherwise it's on the right.
to explain a bit why:
note that $\left(x_{2}-x_{1},y_{2}-y_{1}\right)$ gives the relative position of your point to the line
the scalar product measures the degree of parallelity of some vectors. (it's closely related to the cosine function)
so... do you mean you want to know weather the point $\left(x_{2},y_{2}\right)$ is on the left or on the right of the line passing $\left(x_{1},y_{1}\right)$ and $\left(x_{3},y_{3}\right)$? In this case:
define a normal pointing always to the same side of your line passing the two points $\left(x_{1},y_{1}\right)$ and $\left(x_{3},y_{3}\right)$, that would mean any vector perpendicular on $\left(r_{x},r_{y}\right):=\left(x_{3},y_{3}\right)-\left(x_{1},y_{1}\right)$.
For instance $\left(s_{x},s_{y}\right):=\left(r_{y},-r_{x}\right)$ does it. this vector should point to the left if i'm not wrong :-) you better check this, i might mix up left and right.
then check the sign of the scalar product $\left(s_{x},s_{y}\right)\cdot\left(x_{2}-x_{1},y_{2}-y_{1}\right)$. if it's bigger than $0$ $\left(x_{2},y_{2}\right)$ is on the left, otherwise it's on the right.
to explain a bit why: