Show,by vector method, that the angle between any edge and a face not containing the edge is $\arccos(\frac{1}{\sqrt3})$

Question

Let $k$ be the length of any edge of a regular tetrahedron.Show that the angle between any edge and a face not containing the edge is $\arccos(\frac{1}{\sqrt3})$.

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Let the regular tetrahedron be $OABC$.Let $O$ be the origin and position vectors of $A,B,C$ are the $\vec{a},\vec{b},\vec{c}$.

Let us find the angle between face $OAB$ and the edge $OC$.Angle between a plane and a line is found by finding the angle between the normal to the plane and the line.

The plane $OAB$ is spanned by the vectors $\vec{a}$ and $\vec{b}$.So its normal is given by $\vec{a}\times\vec{b}$.And the edge $OC$ is $\vec{c}$.
Let $\theta$ be the angle between the face $OAB$ and $OC$.So angle between the normal to the face $OAB$ and $OC$ is $\frac{\pi}{2}-\theta$
$\cos(\frac{\pi}{2}-\theta)=\frac{(\vec{a}\times\vec{b}).\vec{c}}{|\vec{a}\times\vec{b}||\vec{c}|}$
$\sin\theta=\frac{(\vec{a}\times\vec{b}).\vec{c}}{|\vec{a}||\vec{b}||\vec{c}|\sin\frac{\pi}{3}}$

I am stuck here and could not solve further.Please help me.Thanks.

Answer 1

Using your idea, let $\theta$ be the angle between the face $OAB$ and $OC$. Then, the angle between the normal $\vec a\times\vec b$ and $OC$ is either $\frac{\pi}{2}\pm\theta$. So, we have

$$\cos\left(\frac{\pi}{2}\pm\theta\right)=\frac{(\vec{a}\times\vec{b})\cdot \vec{c}}{|\vec{a}\times\vec{b}||\vec{c}|}\ ,$$ i.e. $$\mp\sin\theta=\frac{(\vec{a}\times\vec{b})\cdot \vec{c}}{|\vec{a}||\vec{b}||\vec{c}|\sin\frac{\pi}{3}}$$

So, we can have $$|\sin\theta|=\frac{\frac 16|(\vec{a}\times\vec{b})\cdot \vec{c}|}{\frac 16|\vec{a}||\vec{b}||\vec{c}|\sin\frac{\pi}{3}}$$

Since the numerator of the RHS equals the volume of the tetrahedron (why?), we have $$|\sin\theta|=\frac{\frac{\sqrt 2}{12}k^3}{\frac 16k^3\sin\frac{\pi}{3}}=\frac{\sqrt 2}{\sqrt 3}.$$ Then, $\cos\theta=\frac{1}{\sqrt 3}$ follows from this.

Answer 2

First note that $\vec{OC}$ cuts through the midpoint of $AB$ when projected onto $OAB$.

Now, let $\theta$ be the angle between $\vec{OC}$ and the $OAB$. Then,

$$\vec{OC}\cdot \left(\vec{OA} + \vec{OB}\right) = \left|\vec{OC}\right|\left|\vec{OA} + \vec{OB}\right|\cos\theta$$

$$\vec{OC}\cdot\vec{OA} + \vec{OC}\cdot\vec{OB} = (k)(2\cdot \frac{\sqrt{3}}{{2}}k)\cos\theta$$

$$k^2\cos 60^\circ + k^2 \cos 60^\circ = \sqrt{3}k^2\cos\theta$$

$$\frac{1}{2}k^2 + \frac{1}{2}k^2 = \sqrt{3}k^2\cos\theta$$

$$k^2 = \sqrt{3}k^2\cos\theta$$

$$\cos\theta = \frac{1}{\sqrt{3}}$$