Find the angle between the diagonal of a cube and the diagon

[Laurenmath]

Find the angle between the diagonal of a cube and the diagonal of one of its sides.

 

[galactus]

The diagonal of a cube is given by \(\displaystyle \L\\\sqrt{3a^{2}}=\sqrt{3}a\)

 

[pka]

The length of a diagonal on a face is \(\displaystyle a\sqrt 2\) and the length of a diagonal of the cube is \(\displaystyle a\sqrt 3\). Now to find the angle between them apply the law of cosines:
\(\displaystyle a^2 = \left( {a\sqrt 2 } \right)^2 + \left( {a\sqrt 3 } \right)^2 - 2\left( {a\sqrt 2 } \right)\left( {a\sqrt 3 } \right)\cos \left( \theta \right).\)

Solve for \(\displaystyle theta\)

 

[soroban]

Re: Find the angle between the diagonal of a cube and the di

Hello, Laurenmath!

Find the angle between the diagonal of a cube and the diagonal of one of its sides.

If you are familiar with vectors, place the unit cube at the origin.

A diagonal of the cube is: \(\displaystyle \,\vec{u}\:=\:\langle1,1,1\rangle\)
An adjacent diagonal of a face is: \(\displaystyle \,\vec{v}\:=\:\langle1,1,0\rangle\)

The angle between two vectors is given by: \(\displaystyle \L\,\cos\theta\:=\:\frac{\vec{u}\bullet\vec{v}}{|\vec{u}||\vec{v}| }\)

So we have: \(\displaystyle \L\,\cos\theta \:=\:\frac{\langle1,1,1\rangle\bullet\langle1,1,0\rangle}{\sqrt{1^2+1^2+1^2}\cdot{\sqrt{1^2+1^2+0^2}}} \:=\:\frac{1\,+\,1\,+\,0}{\sqrt{3}\cdot\sqrt{2}}\)

Therefore: \(\displaystyle \L\cos\theta\:=\:\frac{2}{\sqrt{6}}\;\;\Rightarrow\;\;\theta \:\approx\:35.26^o\)