How do I find the length from a tetrahedron vertex to its face using vector methods? Suppose the length of each edge of the regular tetrahedron is $x$ and a, b, c represents the position vectors from the origin O and the points A, B, C. How can I find the distance from a vertex to its opposite face?
I thought I had to use the concept of scalar products etc. to do this question, but my solution does not involve any vectors and is largely geometric. The answer does not match.
My thought process:


*

*Since it's a regular tetrahedron, all sides are equilateral triangles.

*I divided the base of the tetrahedron (one triangle) into 3 areas using the centroid of the face to obtain 3 isosceles triangles.

*Let distance from vertex to centroid of on triangular face of the tetrahedron by $k$. Since I know the length of the longest side of the isosceles triangle is $x$, I use sine rule of $\frac{\sin(30)}{k} = \frac{\sin(120)}{x}$ to get my $k$ value in terms of $x$.

*Now I have a new triangle with hypotenuse $x$ and base $k$. I use Pythagoras theorem to get my answer.
Could someone show me where my thought process has gone wrong or show me the direction I should approach this question instead?
 A: Let's use an orthogonal reference, with $O$ at the origin, $\mathbf a$ along the first axis (unit vector $\hat i$), $\mathbf b$ in the  $\hat i\hat j$ plane, and $\mathbf c$ with components along all three axes. Since we know the angle between vectors is $\pi/3$, we can write 
$$\mathbf a=x\hat i\\
\mathbf b=x\cos\frac{\pi}{3}\hat i+x\sin\frac{\pi}{3}\hat j\\
\mathbf c=\alpha \hat i+ \beta \hat j+\gamma\hat k
$$
The distance from $C$ to the plane $ABO$ is $\gamma$. We can now use $$\mathbf a\cdot\mathbf c=x^2\cos\frac{\pi}{3}\\ \mathbf b\cdot\mathbf c=x^2\cos\frac{\pi}{3}\\ \mathbf c\cdot\mathbf c=x^2$$ to calculate all the components.
From the first equation $x\alpha=x^2\cos\frac{\pi}{3}$ or $$\alpha=x/2$$ From the second equation $$x\frac{1}{2}\alpha+x\frac{\sqrt 3}{2}\beta=x^2\frac{1}{2}$$ This yields $$\beta=\frac{x}{2\sqrt 3}$$
The third equation is $$\alpha^2+\beta^2+\gamma^2=x^2$$
Plugging in the previous values for $\alpha$ and $\beta$ we get $$\gamma=x\sqrt\frac{2}{3}$$
A: Since you’re supposed to use “vector methods,” here’s a way to approach this. The vector $\mathbf n=\mathbf a\times\mathbf b$ is orthogonal to the face $\triangle{OAB}$. The distance of $C$ from this face is then the length of the projection of $\mathbf c$ onto $\mathbf n$, namely, $${|\mathbf n\cdot\mathbf c|\over\|\mathbf n\|}={|\mathbf a \times\mathbf b\cdot\mathbf c| \over \|\mathbf a\times\mathbf b\|}.$$
A: Put the tetrahedron in four-dimensional space, with vertices at the points $ (1,0,0,0),(0,1,0,0),(0,0,1,0)$ and $(0,0,0,1)$.  The edges are $\sqrt{2}$.  The centre of one side is $(1/3,1/3,1/3,0)$, and its distance from $(0,0,0,1)$ is $\sqrt{4/3}$.  So your answer is $\sqrt{2/3}$
