In tetrahedron ABCD, AB, AC and AD are perpendicular to each other.
Let:
- $\overrightarrow{AB} = \underline{u}$
- $\overrightarrow{AC} = \underline{v}$
- $\overrightarrow{AD} = \underline{w}$
Given:
- $|AB| = \sqrt 3$
- $|AC| = \sqrt 2$
- $|AD| = \sqrt 6$
The point $X$ is on side $BCD$ such that the vector $$\overrightarrow{AX} = a\underline{u} + b\underline{v} + c\underline{w}$$
is perpendicular to the plane of the side $BCD$.
I know I can use the fact that $\underline{u}^2 = |u|^2$ (with each vector), that $\underline{u} \cdot \underline{v} = 0$ and the same with each length and perpendicular vectors. Moreover, I know that because B, C and D are on the side BCD and X is on BCD, thus I can say that $a+b+c=1$.
From here I don't know how to continue and I would ask for some help (preferably hints or something I have missed / forgot)
Hint: Write $\underline x:=\overrightarrow{AX}$.
Then $\underline x$ is perpendicular to any vector in the plane of the side $BCD$. In particular, it's perpendicular to the vector $\overrightarrow{XB} = \underline u - \underline x$, which means $$\underline x \cdot (\underline u - \underline x )= 0\quad\Longleftrightarrow\quad \underline x\cdot\underline u = \underline x\cdot \underline x .\tag1$$ You can get two additional statements similar to (1) for the other vectors $\overrightarrow{XD}$ and $\overrightarrow{XC}$. This should be enough to solve for $a, b, c$.