# Distance from $A$ to $\triangle BCD$ if $\angle BAC=\angle CAD=\angle DAB=60^\circ$?

Let $$A$$, $$B$$, $$C$$, and $$D$$ be four points in space such that $$\angle BAC=\angle CAD=\angle DAB=60^\circ.$$ If $$AB=1$$, $$AC=2$$, and $$AD=6$$, then what is the distance between $$A$$ and the plane of $$\triangle BCD$$?

Observations: $$ABC$$ and $$BCD$$ are right, as $$ABC$$ is $$30-60-90$$ and then using Law of Cosines we han find the other side lengths for $$BCD.$$ I don't know how to use this to solve for the distance desired though. I have tried letting this distance be $$h$$ and setting up equations.

• One idea would be to construct these points by hand. That's not as awful as it may sound: Without loss of generality, you can pick $A=(0,0,0), B=(1,0,0), C=(1,\sqrt{3},0)$. It then remains to find $D$. Apr 2, 2021 at 20:57

You can use the formula for tetrahedron volume (wiki),

$$V = \frac{abc}{6} \sqrt{1+2 \cos \alpha \cos \beta \cos \gamma - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma}$$

$$\alpha, \beta, \gamma$$ are angles between edges at a vertex and $$a, b, c$$ are lengths of edges from the vertex.

Here we know that at vertex $$A$$, $$\angle BAC = \angle BAD = \angle CAD = 60^0$$ and $$AB = 1, AC = 2, AD = 6$$.

So $$V = \displaystyle \frac{1\cdot2\cdot6}{6} \sqrt{1 + \frac{2}{2^3} - \frac{3}{2^2}} = \sqrt2$$

We also know that the volume of Tetrahedron is given by $$V = \frac{1}{3} A \cdot h$$ where $$A$$ is the area of the base and $$h$$ is the altitude.

As you found out, $$\triangle BCD$$ is right angled triangle with $$BC = \sqrt3, CD = \sqrt{28}, BD = \sqrt{31}$$. So, $$A = \sqrt{21}$$.

So altitude from $$A$$ to base BCD, $$h = \frac{3V}{A} = \sqrt{\frac{6}{7}}$$

Let $$\theta$$ be the angle between the planes $$ABC$$ and $$BCD$$. Recognize the right triangles $$ABC$$ and $$BCD$$ to establish the distance equation

$$AD^2 = (AB+ CD \cos\theta)^2 + BC^2 +(CD\sin\theta)^2$$

which leads to $$\cos\theta =\frac1{\sqrt7}$$. Then, the distance from $$A$$ to the plane $$BCD$$ is $$d= AB \sin\theta = \sqrt{\frac67}$$

• Your solution makes sense, but is there a solution without trigonometry? This question appeared on a test that did not require trigonometry, or memorization of any volume formula. Thanks!
– user797346
Apr 2, 2021 at 22:14
• I realized that, I was referring to the other solutions. However, is it possible to remove trigonometry?
– user797346
Apr 2, 2021 at 22:27
• @WWesEEE - you may replace $a = CD\cos\theta$, $b = CD\sin\theta$, where $a$ is the height of D and $b$ is the horizontal distance between C and D. Note $a^2+b^2 = CD^2=28$ Apr 2, 2021 at 22:36

You can find the volume of the tetrahedron from the lengths of the edges from this Heron-like formula: $${\displaystyle V={\frac {\sqrt {4a^{2}b^{2}c^{2}-a^{2}X^{2}-b^{2}Y^{2}-c^{2}Z^{2}+XYZ}}{12}}}$$ where $$a=AB$$, $$b=AC$$, $$c=AD$$, $$x=CD$$, $$y=BD$$, $$z=BC$$ and {\displaystyle {\begin{aligned}X&=b^{2}+c^{2}-x^{2},\\Y&=a^{2}+c^{2}-y^{2},\\Z&=a^{2}+b^{2}-z^{2}.\end{aligned}}}

Just divide the triple of that volume by the area of $$BCD$$ to find the relative height $$AH$$.

I see from comment that you want to avoid trigonometry: so you can follow these steps.

1. The edges from A form the same angle between them. We can take the z axis from A to be symmetrically placed wrt the three edges. Then they project on the xy plane at $$120^\circ$$.

2. Take the three unitary vectors on the xy plane and add the same z component $$h$$. Take the dot product and determine $$h$$ for the angle to be $$60^\circ$$.

3. Determined $$h$$ you have the three unit vectors along the edges. Multiply by $$1,2,6$$ and get the vertices $$B,C,D$$.

4. Find the plane through $$B,C,D$$ and the distance of $$A$$ from it, or work it out through the determinant of the base (area), and the determinant of the tetrahedron (volume).

Start with a regular tetrahedron AEFD of edge length 6. Choose B on AE and C on AF at the correct distance. As the ratio AC:AB is 2, BC is perpendicular to AE. D dropped perpendicular to AEF cuts the height over AE of triangle AEF 2:1, the same ratio as C. In other words D lies in the plane through C perpendicular to the height which is parallel to BC, hence DC is perpendicular to BC. The volume of ABCD is 1/18 that of AEFD since that is the ratio of areas ABC and AEF. It is also 1/6 of the product BC x CD x d where d is the distance we are looking for. The volume of regular tetrahedron AEFD is $$18 \times \sqrt{2}$$, The lengths BC and CD are (here we need a little bit of Pythagoras) $$\sqrt{3}$$ and $$2 \times \sqrt{7}$$. Therefore d = $$\sqrt{6/7}$$.