Chapter 3, Problem 1P

To determine

The unit vector in the direction of $A$.

Answer to Problem 1P

The unit vector in the direction of $A$ is $\underset{\_}{\left(\widehat{x}0.32+\widehat{z}0.95\right)}$.

Explanation of Solution

Given data:

The first point is $\left(1,-1,-3\right)$.

The second point is $\left(2,-1,0\right)$.

Calculation:

Let the first point is $A_1$ and the second point is $A_2$.

The position vector of point $\left(1,-1,-3\right)$ is given as,

$\overrightarrow{OP}=\widehat{x}x+\widehat{y}y+\widehat{z}z$ (1)

Here,

$O$ is the origin,

$\overrightarrow{OP}$ is the position vector at point $P$.

Substitute $1$ for $x$, $-1$ for $y$, $-3$ for $z$ and $\overrightarrow{O{A}_1}$ for $\overrightarrow{OP}$ in equation (1).

$\begin{array}{c}\overrightarrow{O{A}_1}=\widehat{x}\left(1\right)+\widehat{y}\left(-1\right)+\widehat{z}\left(-3\right)\\ =\widehat{x}-\widehat{y}-3\widehat{z}\end{array}$

Substitute $2$ for $x$, $-1$ for $y$, $0$ for $z$ and $\overrightarrow{O{A}_2}$ for $\overrightarrow{OP}$ in equation (1).

$\begin{array}{c}\overrightarrow{O{A}_2}=\widehat{x}\left(2\right)+\widehat{y}\left(-1\right)+\widehat{z}\left(0\right)\\ =\widehat{x}2-\widehat{y}\end{array}$

The distance vector between the points $A_1$ and $A_2$ is given as,

$A=\overrightarrow{O{A}_2}-\overrightarrow{O{A}_1}$

Here,

$A$ is the distance vector between the points $A_1$ and $A_2$.

Substitute $\widehat{x}-\widehat{y}-3\widehat{z}$ for $\overrightarrow{O{A}_1}$ and $\widehat{x}2-\widehat{y}$ for $\overrightarrow{O{A}_2}$ in the above expression.

$A=\left(\widehat{x}2-\widehat{y}\right)-\left(\widehat{x}-\widehat{y}-3\widehat{z}\right)$

$A=\left(\widehat{x}+3\widehat{z}\right)$ (2)

The general form of a vector is given as,

$A=\left(\widehat{x}{A}_x+\widehat{y}{A}_y+\widehat{z}{A}_z\right)$ (3)

Compare equation (2) and (3), the values are written as,

$A_x=1$

$A_y=0$

$A_z=3$

The magnitude of the vector $A$ is calculated as,

$A=\sqrt{{\left(A_x\right)}^2+{\left(A_y\right)}^2+{\left(A_z\right)}^2}$

Here,

$A$ is the magnitude of the vector $A$.

Substitute $1$ for $A_x$, $0$ for $A_y$ and $3$ for $A_z$ in the above expression.

$\begin{array}{c}A=\sqrt{{\left(1\right)}^2+{\left(0\right)}^2+{\left(3\right)}^2}\\ =\sqrt{1+9}\\ =\sqrt{10}\end{array}$

The unit vector in the direction of $A$ is given as,

$\widehat{a}=\frac{A}{A}$

Here,

$\widehat{a}$ is the unit vector in the direction of $A$.

Substitute $\left(\widehat{x}+3\widehat{z}\right)$ for $A$ and $\sqrt{10}$ for $A$ in the above expression.

$\begin{array}{c}\widehat{a}=\frac{\left(\widehat{x}+3\widehat{z}\right)}{\sqrt{10}}\\ =\left(\widehat{x}0.32+\widehat{z}0.95\right)\end{array}$

Conclusion:

Therefore, the unit vector in the direction of $A$ is $\underset{\_}{\left(\widehat{x}0.32+\widehat{z}0.95\right)}$.