Chapter 10, Problem 6CT

To determine

To find: The component form and magnitude of the vector $\text{v}$ with the initial point $\left(6,2,-5\right)$ and the terminal point $\left(3,-4,8\right)$ .

Answer to Problem 6CT

The component form and magnitude of the vector $\text{v}$ with initial point $\left(6,2,-5\right)$ and terminal point $\left(3,-4,8\right)$ are $\langle -3,-6,13\rangle$ and $\sqrt{214}$ , respectively.

Explanation of Solution

Given information:

The initial point is $\left(6,2,-5\right)$ and the terminal point is $\left(3,-4,8\right)$ .

Calculation:

The component form of a vector $\text{v}$ with initial point $P\left(p_1,p_2,p_3\right)$ and terminal point $Q\left(q_1,q_2,q_3\right)$ is $\text{v}=\langle q_1-p_1,q_2-p_2,q_3-p_3\rangle$ .

Substitute $6$ , $2$ , $-5$ for $p_1$ , $p_2$ , $p_3$ and $3$ , $-4$ , $8$ for $q_1$ , $q_2$ , $q_3$ in the formula.

  $\begin{array}{c}\text{v}=\langle q_1-p_1,q_2-p_2,q_3-p_3\rangle \\ =\langle 3-6,-4-2,8-\left(-5\right)\rangle \\ =\langle -3,-6,13\rangle \end{array}$

So, the component form of the vector $\text{v}$ with initial point $\left(6,2,-5\right)$ and terminal point $\left(3,-4,8\right)$ is $\langle -3,-6,13\rangle$ .

The magnitude of any vector $\text{v}$ with component form $\text{v}=\langle v_1,v_2,v_3\rangle$ is $\Vert \text{v}\Vert =\sqrt{v_1^2+v_2^2+v_3^2}$ .

Substitute $-3$ , $-6$ , $13$ for $v_1$ , $v_2$ , $v_3$ in the formula.

  $\begin{array}{c}\Vert \text{v}\Vert =\sqrt{v_1^2+v_2^2+v_3^2}\\ =\sqrt{{\left(-3\right)}^2+{\left(-6\right)}^2+{\left(13\right)}^2}\\ =\sqrt{9+36+169}\\ =\sqrt{214}\end{array}$

So, the magnitude of vector $\text{v}$ with initial point $\left(6,2,-5\right)$ and terminal point $\left(3,-4,8\right)$ is $\sqrt{214}$ .

Therefore, the component form and magnitude of the vector $\text{v}$ with initial point $\left(6,2,-5\right)$ and terminal point $\left(3,-4,8\right)$ are $\langle -3,-6,13\rangle$ and $\sqrt{214}$ , respectively.