Chapter 10.2, Problem 77E

To determine

To find: The set of all the points $\left(x,y,z\right)$ such that $\Vert \text{v}\Vert =4$ .

Answer to Problem 77E

The set of all the points $\left(x,y,z\right)$ such that $\Vert \text{v}\Vert =4$ is $\left\{\left(x,y,z\right):{\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2+{\left(z-z_1\right)}^2=4^2\right\}$ .

Explanation of Solution

Given information:

The initial and terminal points of $\text{v}$ are $\left(x_1,y_1,z_1\right)$ and $\left(x,y,z\right)$ respectively and $\Vert \text{v}\Vert =4$ .

Calculation:

Use the formula for component form of vector $\text{v}$ with initial point $\langle p_1,p_2,p_3\rangle$ and terminal point $\langle q_1,q_2,q_3\rangle$ .

  $v=\langle q_1-p_1,q_2-p_2,q_3-p_3\rangle$

Substitute $x_1$ , $y_1$ , $z_1$ for $p_1$ , $p_2$ , $p_3$ and $x$ , $y$ , $z$ for $q_1$ , $q_2$ , $q_3$ in the formula.

  $\begin{array}{c}v=\langle q_1-p_1,q_2-p_2,q_3-p_3\rangle \\ v=\langle x-x_1,y-y_1,z-z_1\rangle \end{array}$

Use the formula for magnitude of a vector.

  $\begin{array}{c}\Vert v\Vert =\sqrt{{\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2+{\left(z-z_1\right)}^2}\\ 4=\sqrt{{\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2+{\left(z-z_1\right)}^2}\\ {\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2+{\left(z-z_1\right)}^2=4^2\end{array}$

This is the equation of sphere with center at $\left(x_1,y_1,z_1\right)$ and radius $4$ . So, the set of all the points $\left(x,y,z\right)$ such that $\Vert \text{v}\Vert =4$ is $\left\{\left(x,y,z\right):{\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2+{\left(z-z_1\right)}^2=4^2\right\}$ .

Therefore, the set of all the points $\left(x,y,z\right)$ such that $\Vert \text{v}\Vert =4$ is $\left\{\left(x,y,z\right):{\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2+{\left(z-z_1\right)}^2=4^2\right\}$ .