I have tree points:
$$A=\pmatrix{1\\-2\\-3} B= \pmatrix{-2\\3\\2} C= \pmatrix{0\\2\\4}$$
Now i have to get the height vector that ends on $AC$. The solution should be:
$$f:x = \pmatrix{-2\\3\\2} + r * \pmatrix{-2.1212...\\1.4848..\\-1.1515...} $$
I understand why the vector starts at $B$ but I don't know how to get the direction vector!
Sorry about the .1212.. .4848.., i think you understand what i mean but i didn't know how to format it! I would be glad if somebody can correct my question! Thanks
You want to perform vector rejection: \begin{align*} \vec v &= \vec{AB} - \frac{\vec{AB} \cdot \vec{AC}}{\vec{AC} \cdot \vec{AC}}\vec{AC} \\ &= \begin{pmatrix} -3 \\ 5 \\ 5 \end{pmatrix} - \frac{\begin{pmatrix} -3 \\ 5 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 4 \\ 7 \end{pmatrix}}{\begin{pmatrix} -1 \\ 4 \\ 7 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 4 \\ 7 \end{pmatrix}}\begin{pmatrix} -1 \\ 4 \\ 7 \end{pmatrix} \\ &= \begin{pmatrix} -3 \\ 5 \\ 5 \end{pmatrix} - \frac{29}{33}\begin{pmatrix} -1 \\ 4 \\ 7 \end{pmatrix} \\ &= \begin{pmatrix} -70/33 \\ 49/33 \\ -38/33 \end{pmatrix} \end{align*}