Chapter 9, Problem 37RP

a

To determine

To find:The angle formed by a diagonal of a cube and a diagonal of a face of the cube

Answer to Problem 37RP

The angle formed by a diagonal of a cube and a diagonal of a face of the cube is ${\cos }^{-1}\left(\frac{2}{6}\right)$

Explanation of Solution

Given information:

A diagonal of a cube and a diagonal of a face of the cube

Formula used:

  ${\left(\text{side}\right)}^2={\left(\text{diagonal}\right)}^2+{\left(\text{diagonalofaface}\right)}^2-2\left(\text{diagonal}\right)\left(\text{diagonal of a face}\right)\cos \vartheta$

Calculation :

If side of the cube be ‘a’, then ${\left(\text{side}\right)}^2={\left(\text{diagonal}\right)}^2+{\left(\text{diagonalofaface}\right)}^2-2\left(\text{diagonal}\right)\left(\text{diagonal of a face}\right)\cos \vartheta$

We know that

Length of diagonal of a cube is $\text{a}\sqrt{3}$ and

Length of a diagonal on a face is $\text{a}\sqrt{2}$

Therefore, we have

  $\begin{array}{l}{\text{a}}^2={\left(\text{a}\sqrt{2}\right)}^2\text{+}{\left(\text{a}\sqrt{3}\right)}^2\text{-2}\left(\text{a}\sqrt{2}\right)\left(\text{a}\sqrt{3}\right)\cos \vartheta \\ \Rightarrow \cos \vartheta =\frac{2}{\sqrt{6}}\\ \Rightarrow \vartheta ={\cos }^{-1}\left(\frac{2}{\sqrt{6}}\right)=35.26°\end{array}$

b

To determine

To calculate:The angle formed by two face diagonal that intersect at a vertex of a cube

Answer to Problem 37RP

The angle formed by two face diagonal that intersect at a vertex of a cube ${\cos }^{-1}\left(\frac{2}{\sqrt{2}.\sqrt{3}}\right)$

Explanation of Solution

Given information:

A diagonal of a cube and a diagonal of a face of the cube

Formula used:

  $\overrightarrow{\text{a }}.\overrightarrow{\text{b }}=\left|\overrightarrow{\text{a }}\right|\left|\overrightarrow{\text{b }}\right|\cos \vartheta$

Calculation :

If, OP = $\langle 1,1,0\rangle$ and K= $\langle 1,1,1\rangle$

  

  $\begin{array}{l}\overrightarrow{\text{a }}.\overrightarrow{\text{b }}=\left|\overrightarrow{\text{a }}\right|\left|\overrightarrow{\text{b }}\right|\cos \vartheta \\ \Rightarrow \cos \vartheta =\frac{2}{\sqrt{2}.\sqrt{3}}\\ \Rightarrow \vartheta ={\cos }^{-1}\left(\frac{2}{\sqrt{2}.\sqrt{3}}\right)=35.26°\end{array}$