Question

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

Short answer

The angle between a diagonal of a cube and one of its edges is \({55^\circ }\).

Step-by-step solution

Formula and concept used to find the angle between a diagonal of a cube and one of its edges

The expression to find\({\rm{a}} \cdot {\rm{b}}\)in terms of\(\theta \).

\({\rm{a}} \cdot {\rm{b}} = |{\rm{a}}||{\rm{b}}|\cos \theta \)

Here,

\(|a|\)is the magnitude of a vector,

\(|{\rm{b}}|\)is the magnitude of\({\bf{b}}\)vector, and

\(\theta \)is the angle between vectors\({\bf{a}}\)and\({\bf{b}}\).

Rearrange the equation.

\(\begin{aligned}{l}\cos \theta &= \frac{{a \cdot b}}{{|a||b|}}\\\theta &= {\cos ^{ - 1}}\left( {\frac{{a \cdot b}}{{|a||b|}}} \right)\end{aligned}\) …… (1)

A minimum of two vectors are required to perform a dot product. The resultant dot product of two vectors is a scalar. So, the dot product is also known as scalar product.

Consider a general expression to find the dot product between two three-dimensional vectors.

\(\begin{aligned}{l}{\rm{a}} \cdot {\rm{b}} &= \left\langle {{a_1},{a_2},{a_2}} \right\rangle \cdot \left\langle {{b_1},{b_2},{b_2}} \right\rangle \\{\rm{a}} \cdot {\rm{b}} &= {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}\end{aligned}\) …… (2)

Consider a general expression to find the magnitude of a three dimensional vector that is \(a = \left\langle {{a_1},{a_2},{a_3}} \right\rangle \).

\(|{\rm{a}}| = \sqrt {a_1^2 + a_2^2 + a_3^2} {\rm{ }}\) …… (3)

Similarly, Consider a general expression to find the magnitude of a three dimensional vector that is \({\rm{b}} = \left\langle {{b_1},{b_2},{b_3}} \right\rangle \).

\(|{\rm{b}}| = \sqrt {b_1^2 + b_2^2 + b_3^2} {\rm{ }}\) …… (4)

Calculation to find the angle between a diagonal of a cube and one of its edges

Consider the unit cube, such that its edges lies along the coordinate axis and at the back left corner is at the origin.

The diagonal of the cube which begins at the origin and ends at \((1,1,1)\) has a vector of \(\langle 1,1,1\rangle \). Consider the vector of the edge which begins at the origin and runs along the \(x\)-axis is \(\langle 1,0,0\rangle \).

Let \({\rm{a}} = \langle 1,1,1\rangle \) and \({\rm{b}} = \langle 1,0,0\rangle \).

In equation (2), substitute 1 for \({a_1},{a_2},{a_3}\) and \({b_1},0\) for \({b_2}\) and\({b_3}\).

\(\begin{aligned}{l}a \cdot b &= (1)(1) + (1)(0) + (1)(0)\\ &= 1 + 0 + 0\\ &= 1\end{aligned}\)

In equation (3), substitute \(1\) for\({a_1}\), \({a_2}\) and\({a_3}\).

\(\begin{aligned}{l}|{\rm{a}}| &= \sqrt {{{(1)}^2} + {{(1)}^2} + {{(1)}^2}} \\ &= \sqrt {1 + 1 + 1} \\ &= \sqrt 3 \end{aligned}\)

In equation (4), substitute \(1\) for \({b_1}\) and \(0\) for \({b_2}\) and\({b_3}\).

\(\begin{aligned}{l}|{\rm{b}}| &= \sqrt {{{(1)}^2} + {{(0)}^2} + {{(0)}^2}} \\ &= \sqrt {1 + 0 + 0} \\ &= \sqrt 1 \\ &= 1\end{aligned}\)

In equation (1), substitute \(1\) for\({\rm{a}} \cdot {\rm{b}}\), \(\sqrt 3 \) for \(|{\rm{a}}|\) and 1 for\(|{\rm{b}}|\).

\(\theta = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 3 (1)}}} \right)\)

\( = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)\)

\( = {\cos ^{ - 1}}(0.577)\)

\( = {54.7^\circ }\)

\(\theta \cong {55^\circ }\)

Thus, the angle between a diagonal of a cube and one of its edges is \({55^\circ }\).