Chapter 1, Problem 20P

a)

To determine

The vectors $P$ and $Q$ are unit vectors and this vectors make an angles ${\theta }_1$ and ${\theta }_2$ with the x axis.

Explanation of Solution

Given:

$P=P_x{a}_x+P_y{a}_y$        (I)

$Q=Q_x{a}_x+Q_y{a}_y$        (II)

Calculation:

Consider the two vectors P and Q.

  $P=P_x{a}_x+P_y{a}_y$        (I)

  $Q=Q_x{a}_x+Q_y{a}_y$        (II)

Represent the vector $P$ as shown in figure (1).

Refer the figure (1) and write the expression for the sin and cosine angle.

  $\begin{array}{l}\sin {\theta }_1=\frac{\text{Opposite side}}{\text{Hypotenuse}}\\ \sin {\theta }_1=\frac{P_y}{\left|P\right|}\\ P_y=\left|P\right|\sin {\theta }_1\end{array}$

Similarly,

  $\begin{array}{l}\cos {\theta }_1=\frac{\text{Adjacent}}{\text{Hypotenuse}}\\ \cos {\theta }_1=\frac{P_x}{\left|P\right|}\\ P_x=\left|P\right|\cos {\theta }_1\end{array}$

Calculate the magnitude of P vector using the figure (1).

  $\left|P\right|=\sqrt{P_x^2+P_y^2}$

Substitute $P_x=\left|P\right|\cos {\theta }_1$ and $P_y=\left|P\right|\sin {\theta }_1$ in equation (I).

  $\text{P}=\left(\left|\text{P}\right|\cos {\theta }_1\right)a_x+\left(\left|\text{P}\right|\sin {\theta }_1\right)a_y$

Calculate the unit vector along P.

  $\begin{array}{l}{a}_p=\frac{P_x{a}_x+P_y{a}_y}{\sqrt{P_x^2+P_y^2}}\\ a_p=\frac{\left(\left|P\right|\cos {\theta }_1\right)a_x+\left(\left|P\right|\sin {\theta }_1\right)a_y}{\left|P\right|}\\ a_p=\frac{\left|P\right|\left(\cos {\theta }_1{a}_x+\sin {\theta }_1{a}_y\right)}{\left|P\right|}\\ a_p=\left(\cos {\theta }_1{a}_x+\sin {\theta }_1{a}_y\right)\end{array}$

Thus, $a_p=\left(\cos {\theta }_1{a}_x+\sin {\theta }_1{a}_y\right)$ hence proved.

Similarly represent the vector $Q$ as shown in figure (1).

Refer the figure (2) and write the expression for the sin and cosine angle.

  $\begin{array}{l}\sin {\theta }_2=\frac{\text{Opposite side}}{\text{Hypotenuse}}\\ \sin {\theta }_2=\frac{Q_y}{\left|Q\right|}\\ Q_y=\left|Q\right|\sin {\theta }_2\end{array}$

Similarly,

  $\begin{array}{l}\cos {\theta }_2=\frac{\text{Adjacent}}{\text{Hypotenuse}}\\ \cos {\theta }_2=\frac{Q_x}{\left|Q\right|}\\ Q_x=\left|Q\right|\cos {\theta }_2\end{array}$

Calculate the magnitude of P vector using the figure (2).

  $\left|Q\right|=\sqrt{Q_x^2+Q_y^2}$

Substitute $Q_x=\left|Q\right|\cos {\theta }_1$ and $Q_y=\left|Q\right|\sin {\theta }_2$ in equation (I).

  $\text{Q}=\left(\left|\text{Q}\right|\cos {\theta }_2\right)a_x+\left(\left|\text{Q}\right|\sin {\theta }_2\right)a_y$

Calculate the unit vector along P.

  $\begin{array}{l}{a}_Q=\frac{Q_x{a}_x+Q_y{a}_y}{\sqrt{Q_x{}^2+Q_y^2}}\\ a_Q=\frac{\left(\left|Q\right|\cos {\theta }_2\right)a_x+\left(\left|Q\right|\sin {\theta }_2\right)a_y}{\left|Q\right|}\\ a_Q=\frac{\left|Q\right|\left(\left(\cos {\theta }_2\right)a_x+\left(\sin {\theta }_2\right)a_y\right)}{\left|Q\right|}\\ a_Q=\left(\cos {\theta }_2\right)a_x+\left(\sin {\theta }_2\right)a_y\end{array}$

Thus, $a_Q=\left(\cos {\theta }_2{a}_x+\sin {\theta }_2{a}_y\right)$ hence proved.

b)

To determine

The formula for $\cos \left({\theta }_2-{\theta }_1\right)$ and $\cos \left({\theta }_2+{\theta }_1\right)$.

Explanation of Solution

Consider P and Q is making an angle of ${\theta }_1\text{ and }{\theta }_{\text{2}}$ respectively with the x axis as shown in figure (3).

Calculate the dot product of P and Q.

  $P\cdot Q=PQ\cos {\theta }_{PQ}$

Here, the smaller angle between P and Q is ${\theta }_{PQ}$.

  $\begin{array}{l}\frac{P·Q}{|P\Vert Q|}=\cos \left({\theta }_2-{\theta }_1\right)\hfill \\ \begin{array}{l}\frac{P·Q}{|P\Vert Q|}=\cos \left({\theta }_2-{\theta }_1\right)\\ \begin{array}{l}\frac{\left(P_x,P_y\right)\left(Q_x,Q_y\right)}{\left|P\right|\left|Q\right|}=\cos \left({\theta }_2-{\theta }_1\right)\hfill \\ \begin{array}{l}\frac{P_z{Q}_x+P_y{Q}_y}{|P\Vert Q|}=\cos \left({\theta }_2-{\theta }_1\right)\\ \begin{array}{l}\frac{P_s{Q}_x+P_y{Q}_x}{\left|P\right|\left|Q\right|}=\cos \left({\theta }_2-{\theta }_1\right)\hfill \\ \frac{P_x}{\left|P\right|}\frac{Q_x}{\left|Q\right|}+\frac{P_y}{\left|P\right|}\frac{Q_y}{\left|Q\right|}=\cos \left({\theta }_2-{\theta }_1\right)\hfill \end{array}\end{array}\hfill \end{array}\end{array}\hfill \end{array}$

Substitute the respective value in the above equation and obtain.

  $\begin{array}{l}\cos \left({\theta }_2-{\theta }_1\right)=\frac{P_x}{\left|P\right|}\frac{Q_x}{\left|Q\right|}+\frac{P_y}{\left|P\right|}\frac{Q_y}{\left|Q\right|}\\ \cos \left({\theta }_2-{\theta }_1\right)=\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2\end{array}$

Thus, the formula for $\cos \left({\theta }_2-{\theta }_1\right)$ is $\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2$.

Consider the vectors P and Q are making angle ${\theta }_1\text{ and }{\theta }_2$ with the x axis as shown in figure (4).

Calculate the dot product of P and Q.

  $P\cdot Q=PQ\cos {\theta }_{PQ}$

Here, the smaller angle between P and Q is ${\theta }_{PQ}={\theta }_1+{\theta }_2$.

  $\begin{array}{l}P·Q=PQ\cos {\theta }_{PQ}\\ \frac{P·Q}{|P\Vert Q|}=\cos \left({\theta }_1+{\theta }_2\right)\\ \frac{P·Q}{|P\Vert Q|}=\cos \left({\theta }_1+{\theta }_2\right)\end{array}$

  $\begin{array}{l}\frac{\left(P_x,P_y\right)\left(Q_x,-Q_y\right)}{\left|P\right|\left|Q\right|}=\cos \left({\theta }_1+{\theta }_2\right)\\ \frac{P_x{Q}_x-P_y{Q}_y}{|P\Vert Q|}=\cos \left({\theta }_1+{\theta }_2\right)\\ \frac{P_x{Q}_x-P_y{Q}_y}{|P\Vert Q|}=\cos \left({\theta }_1+{\theta }_2\right)\end{array}$

  $\frac{P_x{Q}_x}{|P\Vert Q|}-\frac{P_y{Q}_y}{|P\Vert Q|}=\cos \left({\theta }_1+{\theta }_2\right)$

Substituting sin and cosine value in the above equation and simplified as shown below.

  $\begin{array}{l}\cos \left({\theta }_1+{\theta }_2\right)=\frac{P_x}{\left|P\right|}\frac{Q_x}{\left|Q\right|}-\frac{P_y}{\left|P\right|}\frac{Q_y}{\left|Q\right|}\\ \cos \left({\theta }_1+{\theta }_2\right)=\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_1\sin {\theta }_2\end{array}$

Thus, the formula for $\cos \left({\theta }_1+{\theta }_2\right)$ is $\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_1\sin {\theta }_2$.

c)

To determine

The expression for $\frac{1}{2}\left|P-Q\right|$ if P is making an angle $\theta$ with Q.

Explanation of Solution

Consider the two vectors P and Q making an angle ${\theta }_1\text{ and }{\theta }_2$ with the x axis.

Calculate the $P-Q$.

  $P-Q=\left(\cos {\theta }_1-\cos {\theta }_2\right)a_x+\left(\sin {\theta }_1-\sin {\theta }_2\right)a_y$

Calculate the magnitude of $\left|P-Q\right|$.

  $\begin{array}{c}\left|P-Q\right|=\sqrt{{\left(\cos {\theta }_1-\cos {\theta }_2\right)}^2+{\left(\sin {\theta }_1-\sin {\theta }_2\right)}^2}\\ =\sqrt{{\cos }^2{\theta }_1+{\cos }^2{\theta }_2-2\cos {\theta }_1\cos {\theta }_2+{\sin }^2{\theta }_1+{\sin }^2{\theta }_2-2\sin {\theta }_1\sin {\theta }_2}\\ =\sqrt{{\cos }^2{\theta }_1+{\sin }^2{\theta }_1+{\cos }^2{\theta }_2+{\sin }^2{\theta }_2-2\cos {\theta }_1\cos {\theta }_2-2\sin {\theta }_1\sin {\theta }_2}\\ =\sqrt{2-2\left(\cos {\theta }_1\cos {\theta }_2+\sin \theta \sin {\theta }_2\right)}\end{array}$

Calculate the $\frac{1}{2}\left|P-Q\right|$.

  $\begin{array}{c}\frac{1}{2}\left|P-Q\right|==\frac{1}{2}\sqrt{2\left(1-\cos \left({\theta }_2-{\theta }_1\right)\right)}\\ =\sqrt{\frac{\left(1-\cos \left({\theta }_2-{\theta }_1\right)\right)}{2}}\\ =\sqrt{{\sin }^2\frac{\left({\theta }_2-{\theta }_1\right)}{2}}\\ =\left|\sin \frac{\left({\theta }_2-{\theta }_1\right)}{2}\right|\end{array}$

Thus, $\frac{1}{2}\left|P-Q\right|$ is equal to in terms of $\theta$ is $\left|\sin \frac{\left({\theta }_2-{\theta }_1\right)}{2}\right|$.