Chapter 1, Problem 60P

(a)

To determine

The unit vector in the northeast direction.

Answer to Problem 60P

The unit vector in the northeast direction is $\frac{1}{\sqrt{2}}\widehat{i}+\frac{1}{\sqrt{2}}\widehat{j}$.

Explanation of Solution

Given:

Unit vectors $\widehat{i}$ and $\widehat{j}$ are directed in east and north direction respectively.

Formula used:

The magnitude of unit vector is always equal to one that’s why it is termed as a unit vector.

The general expression for a vector is $\overrightarrow{v}=v_x\widehat{i}+v_y\widehat{j}$ .

Write the expression for vector in terms of its components.

  $\overrightarrow{v}=v\cos \theta \widehat{i}+v\sin \theta \widehat{j}$   ............... (1)

Here, $\overrightarrow{v}$ is the vector and $v$ is the magnitude of vector.

When the unit vector is in the northeast direction, it will have an angle $45°$ with $+x$ axis.

Calculation:

Substitute $1$ for $v$ and $45°$ for $\theta$ in equation (1).

  $\begin{array}{c}\overrightarrow{v}=1\cos 45°\widehat{i}+1\sin 45°\widehat{j}\\ =\frac{1}{\sqrt{2}}\widehat{i}+\frac{1}{\sqrt{2}}\widehat{j}\end{array}$

Conclusion:

Thus, the unit vector in the northeast direction is $\frac{1}{\sqrt{2}}\widehat{i}+\frac{1}{\sqrt{2}}\widehat{j}$.

(b)

To determine

The unit vector in $70°$ clockwise from $-y$ axis.

Answer to Problem 60P

The unit vector in $70°$ clockwise from $-y$ axis is $-\frac{\sqrt{3}}{2}\widehat{i}-\frac{1}{2}\widehat{j}$ .

Explanation of Solution

Given:

Unit vectors $\widehat{i}$ and $\widehat{j}$ are directed in east and north direction respectively.

Formula used:

The magnitude of unit vector is always equal to one that’s why it is termed as a unit vector.

The general expression for a vector is $\overrightarrow{v}=v_x\widehat{i}+v_y\widehat{j}$ .

Write the expression for vector in terms of its components.

  $\overrightarrow{v}=v\cos \theta \widehat{i}+v\sin \theta \widehat{j}$   ............... (1)

Here, $\overrightarrow{v}$ is the vector and $v$ is the magnitude of vector.

When the unit vector is in $70°$ clockwise from $-y$ axis it will make $30°$ with $-x$ axis.

Calculation:

Substitute $-1$ for $v$ and $30°$ for $\theta$ in equation (1).

  $\begin{array}{c}\overrightarrow{v}=-\cos 30°\widehat{i}-1\sin 30°\widehat{j}\\ =-\frac{\sqrt{3}}{2}\widehat{i}-\frac{1}{2}\widehat{j}\end{array}$

Conclusion:

Thus,the unit vector in $70°$ clockwise from $-y$ axis. $-\frac{\sqrt{3}}{2}\widehat{i}-\frac{1}{2}\widehat{j}$.

To determine

The unit vector in the southwest direction.

Answer to Problem 60P

The unit vector in the southwest direction is $-\frac{1}{\sqrt{2}}\widehat{i}-\frac{1}{\sqrt{2}}\widehat{j}$ .

Explanation of Solution

Given:

Unit vectors $\widehat{i}$ and $\widehat{j}$ are directed in east and north direction respectively.

Formula used:

Magnitude of unit vector is always equal to one that’s why it is termed as a unit vector.

The general expression for a vector is $\overrightarrow{v}=v_x\widehat{i}+v_y\widehat{j}$ .

Write the expression for vector in terms of its components.

  $\overrightarrow{v}=v\cos \theta \widehat{i}+v\sin \theta \widehat{j}$   ............... (1)

Here, $\overrightarrow{v}$ is the vector and $v$ is the magnitude of vector.

When the unit vector is in the southwest direction, it will have an angle $45°$ with $-x$ axis.

Calculation:

Substitute $-1$ for $v$ and $45°$ for $\theta$ in equation (1).

  $\begin{array}{c}\overrightarrow{v}=-1\cos 45°\widehat{i}-1\sin 45°\widehat{j}\\ =-\frac{1}{\sqrt{2}}\widehat{i}-\frac{1}{\sqrt{2}}\widehat{j}\end{array}$

Conclusion:

Thus, the unit vector in the southwest direction is $-\frac{1}{\sqrt{2}}\widehat{i}-\frac{1}{\sqrt{2}}\widehat{j}$.