Chapter 9.4, Problem 48E

(a)

To determine

To proof: The tetrahedron of vertices $A\left(1,0,0\right)$, $B\left(0,1,0\right)$, $C\left(0,0,1\right)$ and $D\left(1,1,1\right)$ is a regular tetrahedron.

Explanation of Solution

Proof:

A tetrahedron is said to be regular when all sides are of same length.

The vertices of the tetrahedron are $A\left(1,0,0\right)$, $B\left(0,1,0\right)$, $C\left(0,0,1\right)$ and $D\left(1,1,1\right)$.

The distance between two points $P\left(x_1,y_1,z_1\right)$ and $Q\left(x_2,y_2,z_2\right)$ is,

$D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$ (1)

Where,

• $D$ is the distance between two points.

Substitute $1$ for $x_2$, $1$ for $x_1$, $1$ for $y_2$, $0$ for $y_1$, $1$ for $z_2$, $0$ for $z_1$ and $DA$ for $D$ in equation (1) to find $DA$.

$\begin{array}{c}DA=\sqrt{{\left(1-1\right)}^2+{\left(1-0\right)}^2+{\left(1-0\right)}^2}\\ =\sqrt{2}\end{array}$

The length of the sides $DA$ is $\sqrt{2}$.

Substitute $1$ for $x_2$, $0$ for $x_1$, $1$ for $y_2$, $1$ for $y_1$, $1$ for $z_2$, $0$ for $z_1$ and $DB$ for $D$ in equation (1) to find $DB$.

$\begin{array}{c}DB=\sqrt{{\left(1-0\right)}^2+{\left(1-1\right)}^2+{\left(1-0\right)}^2}\\ =\sqrt{2}\end{array}$

The length of the sides $DB$ is $\sqrt{2}$.

Substitute $1$ for $x_2$, $0$ for $x_1$, $1$ for $y_2$, $0$ for $y_1$, $1$ for $z_2$, $1$ for $z_1$ and $DC$ for $D$ in equation (1) to find $DC$.

$\begin{array}{c}DC=\sqrt{{\left(1-0\right)}^2+{\left(1-0\right)}^2+{\left(1-1\right)}^2}\\ =\sqrt{2}\end{array}$

The length of the sides $DC$ is $\sqrt{2}$.

Substitute $0$ for $x_2$, $1$ for $x_1$, $0$ for $y_2$, $0$ for $y_1$, $1$ for $z_2$, $0$ for $z_1$ and $AC$ for $D$ in equation (1) to find $AC$.

$\begin{array}{c}AC=\sqrt{{\left(0-1\right)}^2+{\left(0-0\right)}^2+{\left(1-0\right)}^2}\\ =\sqrt{2}\end{array}$

The length of the sides $AC$ is $\sqrt{2}$.

Substitute $0$ for $x_2$, $1$ for $x_1$, $1$ for $y_2$, $0$ for $y_1$, $0$ for $z_2$, $0$ for $z_1$ and $AB$ for $D$ in equation (1) to find $AB$.

$\begin{array}{c}AB=\sqrt{{\left(0-1\right)}^2+{\left(1-0\right)}^2+{\left(0-0\right)}^2}\\ =\sqrt{2}\end{array}$

The length of the sides $AB$ is $\sqrt{2}$.

Substitute $0$ for $x_2$, $0$ for $x_1$, $0$ for $y_2$, $1$ for $y_1$, $1$ for $z_2$, $0$ for $z_1$ and $BC$ for $D$ in equation (1) to find $BC$.

$\begin{array}{c}BC=\sqrt{{\left(0-0\right)}^2+{\left(0-1\right)}^2+{\left(1-0\right)}^2}\\ =\sqrt{2}\end{array}$

The length of the sides $BC$ is $\sqrt{2}$.

Hence, all the sides of tetrahedron is equal it is regular tetrahedron.

(b)

To determine

The angle between the vector that join the centre to any two of the vertices.

Answer to Problem 48E

The angle between the vector that join the centre to any two of the vertices is $109.47°$.

Explanation of Solution

Given:

The vertices of the tetrahedron are $A\left(1,0,0\right)$, $B\left(0,1,0\right)$, $C\left(0,0,1\right)$ and $D\left(1,1,1\right)$.

The centre of the tetrahedron is $E\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$.

Calculation:

The vector that determines the two points $P\left(x_1,y_1,z_1\right)$ and $Q\left(x_2,y_2,z_2\right)$ is,

$\overrightarrow{PQ}=\left(x_2-x_1\right)i\text{+}\left(y_2-y_1\right)j\text{+}\left(z_2-z_1\right)k$ (1)

The vector $\overrightarrow{EB}$ is,

Substitute $0$ for $x_2$, $\frac{1}{2}$ for $x_1$, $1$ for $y_2$, $\frac{1}{2}$ for $y_1$, $0$ for $z_2$, $\frac{1}{2}$ for $z_1$ and $\overrightarrow{EB}$ for $\overrightarrow{PQ}$ in equation (1) to find $\overrightarrow{EB}$.

$\begin{array}{c}\overrightarrow{EB}=\left(0-\frac{1}{2}\right)i\text{+}\left(1-\frac{1}{2}\right)j\text{+}\left(0-\frac{1}{2}\right)k\\ \text{=}-\frac{1}{2}i\text{+}\frac{1}{2}j-\frac{1}{2}k\end{array}$

The vector $\overrightarrow{EA}$ is,

Substitute $1$ for $x_2$, $\frac{1}{2}$ for $x_1$, $0$ for $y_2$, $\frac{1}{2}$ for $y_1$, $0$ for $z_2$, $\frac{1}{2}$ for $z_1$ and $\overrightarrow{EA}$ for $\overrightarrow{PQ}$ in equation (1) to find $\overrightarrow{EA}$.

$\begin{array}{c}\overrightarrow{EA}=\left(1-\frac{1}{2}\right)i\text{+}\left(0-\frac{1}{2}\right)j\text{+}\left(0-\frac{1}{2}\right)k\\ \text{=}\frac{1}{2}i-\frac{1}{2}j-\frac{1}{2}k\end{array}$

The dot product of vector $\overrightarrow{EB}$ and $\overrightarrow{EA}$ is,

$\begin{array}{c}\overrightarrow{EA}\cdot \overrightarrow{EA}=\left(-\frac{1}{2}i\text{+}\frac{1}{2}j-\frac{1}{2}k\right)\left(\frac{1}{2}i-\frac{1}{2}j-\frac{1}{2}k\right)\\ =\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)\\ =-\frac{1}{4}\end{array}$

The magnitude of the vector $\overrightarrow{EB}$ is,

$\begin{array}{c}\left|\overrightarrow{EB}\right|=\sqrt{{\left(-\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2+{\left(-\frac{1}{2}\right)}^2}\\ =\frac{\sqrt{3}}{2}\end{array}$

The magnitude of the vector $\overrightarrow{EA}$ is,

$\begin{array}{c}\left|\overrightarrow{EA}\right|=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(-\frac{1}{2}\right)}^2+{\left(-\frac{1}{2}\right)}^2}\\ =\frac{\sqrt{3}}{2}\end{array}$

Formula to calculate the angle between two vector is,

$\theta ={\cos }^{-1}\left(\frac{\overrightarrow{EB}\cdot \overrightarrow{EA}}{\left|\overrightarrow{EB}\right|\left|\overrightarrow{EA}\right|}\right)$

Where,

• $\theta$ is the angle between two vector.

Substitute $-\frac{1}{4}$ for $\overrightarrow{EB}\cdot \overrightarrow{EA}$, $\frac{\sqrt{3}}{2}$ for $\left|\overrightarrow{EB}\right|$ and $\frac{\sqrt{3}}{2}$ for $\overrightarrow{EA}$ to find $\theta$.

$\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{-\frac{1}{4}}{\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}}\right)\\ =109.47°\end{array}$

Thus, the angle between the vector that join the centre to any two of the vertices is $109.47°$.