Given two vectors:

\[ u = \begin{vmatrix} u1 \\ u2 \end{vmatrix} \]

\[ v = \begin{vmatrix} v1 \\ v2 \end{vmatrix} \]

if we project a perpendicular line from (v1, v2) to (u1, u2), meeting at a point with length p from the origin, prove that:

\[\begin{align*} u \cdot v & = p \cdot ||u|| \\ \text{where } u \cdot v & = u_{1} v_{1} + u_{2} v_{2} \text{    (i.e. the inner product)} \end{align*} \]

Let α and β be the respective angles of v and u from the x-axis:

\[\begin{aligned} p & = ||v|| \space cos (\alpha - \beta) \\ p & = ||v|| \space (cos \space \alpha \space cos \space \beta + sin \space \alpha \space \sin \space \beta) \\ p \cdot ||u|| & = ||v|| \cdot ||u|| \space (cos \space \alpha \space cos \space \beta + sin \space \alpha \space \sin \space \beta) \\ p \cdot ||u|| & = ||v|| \space cos \space \alpha \cdot ||u|| \space cos \space \beta + ||v|| \space sin \space \alpha \cdot ||u|| \space \sin \space \beta \\ p \cdot ||u|| & = v_{1} u_{1} + v_{2} u_{2} \\ p \cdot ||u|| & = u \cdot v \end{aligned} \]

# posted by rot13(Unafba Pune) @ 12:26 AM