Let $\vec{p}_0$ and $\vec{p}_1$ be the vertices at the base of the triangle, with $\vec{p}_2$ the vertex at apex.
Define $\hat{b}$ as the base unit vector,
$$\hat{b} = \frac{\vec{p}_1 - \vec{p}_0}{\left\lVert\vec{p}_1 - \vec{p}_0\right\rVert}$$
where $\lVert\vec{p}_1 - \vec{p}_0\rVert = \sqrt{(\vec{p}_1-\vec{p}_0)\cdot(\vec{p}_1-\vec{p}_0)}$.
If we project the vector from first base vertex to apex vertex, $(\vec{p}_2 - \vec{p}_0)$, to the base unit vector $\hat{b}$ (using vector dot product), we get the (signed) distance on the base to a point directly below the apex vertex. The distance between that point and the apex is the height $h$ of the triangle:
$$h = \left\lVert (\vec{p}_2 - \vec{p}_0) - \hat{b} ( \hat{b} \cdot (\vec{p}_2 - \vec{p}_0) \right\rVert$$
Substituting $\hat{b}$ we get
$$h = \left\lVert \vec{p}_2 - \vec{p}_0 - \frac{\left(\vec{p}_1 - \vec{p}_0\right)\left(\left( \vec{p}_1 - \vec{p}_0 \right)\cdot\left(\vec{p}_2 - \vec{p}_0 \right)\right)}{\left(\vec{p}_1 - \vec{p}_0\right)\cdot\left(\vec{p}_1 - \vec{p}_0\right)} \right\rVert$$
In pseudocode, a function that calculates the height given the base vertex coordinates x0,y0,z0 and x1,y1,z1 and the apex vertex coordinates x2,y2,z2, is
function triangle_height(x0,y0,z0, x1,y1,z1, x2,y2,z2):
# x0,y0,z0 First base vertex
# x1,y1,z1 Second base vertex
# x2,y2,z2 Apex vertex
tmp0 = x1 - x0
tmp1 = y1 - y0
tmp2 = z1 - z0
tmp3 = x2 - x0
tmp4 = y2 - y0
tmp5 = z2 - z0
tmp6 = tmp0*tmp0 + tmp1*tmp1 + tmp2*tmp2
if (tmp6 <= 0) then
# Degenerate triangle; (x0,y0,z0) = (x1,y1,z1).
# It is not a triangle, but a line (or a point).
# This returns the length of the line,
# or 0 if (x0,y0,z0) = (x1,y1,z1) = (x2,y2,z2).
hh = tmp3*tmp3 + tmp4*tmp4 + tmp5*tmp5
else
tmp7 = (tmp0*tmp3 + tmp1*tmp4 + tmp2*tmp5) / tmp6
# tmpx,tmpy,tmpz is the apex vector, perpendicular to base
tmpx = tmp3 - tmp7 * tmp0
tmpy = tmp4 - tmp7 * tmp1
tmpz = tmp5 - tmp7 * tmp2
hh = tmpx*tmpx + tmpy*tmpy + tmpz*tmpz
end if
return sqrt(hh)
The above pseudocode is hand-tuned from Maple codegen module output from the vector formula, adding the check. Although math says tmp6 cannot be negative, it is better practice to check if it is nonpositive instead. You see, any floating-point comparison against an exact value is always suspect, even if the target value is zero.
The Wolfram Mathworld article on 3D Point-Line Distance mentions in passing an even simpler formula:
$$h = \frac{\left\lVert\left(\vec{p}_1 - \vec{p}_0\right)\times\left(\vec{p}_0 - \vec{p}_2\right)\right\rVert}{\left\lVert\vec{p}_1 - \vec{p}_0\right\rVert}$$
(Note the different numbering here, $2\gets 0$, $0\gets 1$, $1\gets 2$, compared to the Mathworld page.)
In pseudocode:
function triangle_height(x0,y0,z0, x1,y1,z1, x2,y2,z2):
# x0,y0,z0 First base vertex
# x1,y1,z1 Second base vertex
# x2,y2,z2 Apex vertex
tmp0 = x1 - x0
tmp1 = y1 - y0
tmp2 = z1 - z0
tmp3 = x0 - x2
tmp4 = y0 - y2
tmp5 = z0 - z2
tmp6 = tmp0*tmp0 + tmp1*tmp1 + tmp2*tmp2
if (tmp6 <= 0) then
hh = tmp3*tmp3 + tmp4*tmp4 + tmp5*tmp5
else
tmp7 = tmp5*tmp1 - tmp2*tmp4
tmp8 = tmp2*tmp3 - tmp5*tmp0
tmp9 = tmp4*tmp0 - tmp1*tmp3
hh = (tmp7*tmp7 + tmp8*tmp8 + tmp9*tmp9) / tmp6
end if
return sqrt(hh)
Both functions are mathematically the same (but using floating point numbers, the rounding errors may differ; so do not expect the results to be exactly the same when using floating-point math). I don't see any significant difference between the two. In particular, both need twelve multiplications and one division. The latter does need fewer additions/subtractions in the non-degenerate-triangle cases, so it might be a tiny bit faster.
The equation of a variable plane in intercept form is $$\frac xa+\frac yb+ \frac zc=1$$
Here, $a,b$ and $c$ are intercept on $x,y$ and $z$ axes.
Given, it's distance from origin is constant $$\frac{\Big|\Big(\frac xa+\frac yb+ \frac zc-1\Big)_{x=y=z=0}\Big|}{\sqrt{\frac {1}{a^2}+\frac {1}{b^2}+\frac {1}{c^2}}}=p(\text{constant})$$
And centroid of triangle $$(x,y,z)=\left(\frac a3,\frac b3,\frac c3\right)$$
Therefore you get $$\frac{1}{p^2}=\frac {1}{a^2}+\frac {1}{b^2}+\frac {1}{c^2}$$
Also, $a=3x,b=3y$ and $c=3z$
$$\implies \frac{1}{p^2}=\frac {1}{9x^2}+\frac {1}{9y^2}+\frac {1}{9z^2}$$
Hence the desired locus is :
$$\color{blue}{\frac {1}{x^2}+\frac {1}{y^2}+\frac {1}{z^2}=\frac{9}{p^2}}$$
Best Answer
HINT.
1) The distance between given vertices is not $\sqrt{37}$: please revise your calculation.
2) You are given the sides of the rhombus and one of its diagonals, but it is easy to find the other diagonal (Pythagora's theorem).
3) Once you have both diagonals, you can find the area of the rhombus and then divide by the side to get the altitude.