Question

To find: The angle between a diagonal of a cube and a diagonal of one of its faces.

Short answer

The angle between a diagonal of a cube and a diagonal of one of its faces is \({35^\circ }\).

Step-by-step solution

formula and concept used to find the angle between a diagonal of a cube and a diagonal of one of its faces

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.

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

Minimum 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 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 a diagonal of one of its faces

Consider the unit cube, such that its edges lies along the three coordinate axis and within the first octant.

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 representation of a diagonal with one of its faces is \(\langle 1,1,0\rangle \)

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

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

\(\begin{aligned}{l}{\rm{a}} \cdot {\rm{b}} &= (1)(1) + (1)(1) + (1)(0)\\ &= 1 + 1 + 0\\ &= 2\end{aligned}\)

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\({b_1}\), \({b_2}\) and \(0\) for\({b_3}\).

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

In equation (1), substitute \(2\) for\(a \cdot b\), \(\sqrt 3 \) for \(|a|\)and \(\sqrt 2 \) for \(|b|\).

\(\begin{aligned}{l}\theta &= {\cos ^{ - 1}}\left( {\frac{2}{{\sqrt 3 \sqrt 2 }}} \right)\\ &= {\cos ^{ - 1}}(0.816)\\ &= {35^\circ }\end{aligned}\)

Thus, the angle between a diagonal of a cube and a diagonal of one of its faces is \(\underline {{{35}^\circ }} \).