Find the vector $\vec{c}$ which is orthogonal to $\vec{a}$ and $\vec{b}$ and whose first component is 1
$\vec{a}$ = (0 / 4 / -2), start point: $P$ (2 / -3 / 5), end point: $A$ (2 / 1 / 3)
$\vec{b}$ = (-5 / 5 / -1), start point: $P$ (2 / -3 / 5), end point: $B$ (-3 / 2 / 4)
I can only assume that $\vec{c}$ is supposed to be an unit vector. What I don't understand, is, how one vector is supposed to be orthogonal to both vectors $\vec{a}$ and $\vec{b}$ at once?
Maybe the Point $P$ provides the place for the new vector to be orthogonal to both vectors, but how to find $\vec{c}$ respectively determine its properties (direction, x-, y-, and z-values)?
Edit - As suggested by rschwieb here's the solution I worked out:
The first component of $\vec{c}$ is $x = 1$, the other ones are unknown, thus:
$\vec{c}$ = (1, y, z)
Because the vector $\vec{c}$ is orthogonal to the other vectors $\vec{a}$ and $\vec{b}$, the scalar products are: $\vec{c} * \vec{a} = 0$ and $\vec{c} * \vec{b} = 0$. This results in the following system of equations:
$\vec{a} * \vec{c} => (0, 4, -2) * (1, y, z) = (0*1) + (4*y) + (-2*z) = 0$
$\vec{b} * \vec{c} => (-5, 5, -1) * (1, y, z) = (-5*1) + (5*y) + (-1*z) = 0$
Which is then simplified to:
$4y - 2z = 0$
$-5 + 5y - z = 0$
When calculating this system of equations, the solutions then are
y = $\dfrac{5}{3}$ and z = $\dfrac{10}{3}$
Thus, the end result is:
$\vec{c} = (1, \dfrac{5}{3}, \dfrac{10}{3})$