Chapter 1, Problem 1P

Determine the angles made by the vector $V=-36i+15j$ with the positive x- and y-axes. Write the unit vector n in the direction of V.

To determine

The angle made by the given vector with the positive $x-$ and $y-$ axes.

The unit vector $n$ in the direction of $V$.

Answer to Problem 1P

The angle made by the given vector with the positive $x-$ is $157.38°$ and $y-$ axes is $67.38°$.

The unit vector $n$ in the direction of $V$ is $-0.923i+0.385j$.

Explanation of Solution

Given information:

The given vector is $V=-36i+15j$.

Write the expression for the magnitude of vector.

$\left|V\right|=\sqrt{a^2+b^2}$   ........(I)

Here, the magnitude of vector is $\left|V\right|$, the component of vector in $x-$ direction is $a$, and the component of vector in $y-$ direction is $b$.

Write the expression for angle made by vector with positive $x-$ axis.

$\cos {\theta }_{\text{x}}=\frac{a}{\left|V\right|}$   ........(II)

Here, the angle made by vector with positive $x-$ axis. is ${\theta }_{\text{x}}$.

Write the expression for angle made by vector with positive $y-$ axis.

$\cos {\theta }_{\text{y}}=\frac{b}{\left|V\right|}$   ........(II)

Here, the angle made by vector with positive $y-$ axis. is ${\theta }_{\text{y}}$.

Write the expression for the unit vector.

$n=\frac{V}{\left|V\right|}$   ........(III)

Here, the unit vector in the direction of $V$ is $n$.

Calculation:

Substitute $-36$ for $a$ and $15$ for $b$ in Equation (I).

$\begin{array}{c}\left|V\right|=\sqrt{{\left(-36\right)}^2+{\left(15\right)}^2}\\ =\sqrt{1296+225}\\ =39\end{array}$

Substitute $-36$ for $a$ and $39$ for $\left|V\right|$ in Equation (II).

$\begin{array}{c}\cos {\theta }_{\text{x}}=\frac{-36}{39}\\ {\theta }_{\text{x}}={\cos }^{-1}\left(\frac{-36}{39}\right)\\ =157.38°\end{array}$

Substitute $15$ for $b$ and $39$ for $\left|V\right|$ in Equation (III).

$\begin{array}{c}\cos {\theta }_{\text{y}}=\frac{15}{39}\\ {\theta }_{\text{y}}={\cos }^{-1}\left(\frac{15}{39}\right)\\ =67.38°\end{array}$

Substitute $-36i+15j$ for $V$ and $39$ for $\left|V\right|$ in Equation (IV).

$\begin{array}{c}n=\frac{-36i+15j}{39}\\ =-0.923i+0.385j\end{array}$

Conclusion:

The angle made by the given vector with the positive $x-$ is $157.38°$ and $y-$ axes is $67.38°$.

The unit vector $n$ in the direction of $V$ is $-0.923i+0.385j$.