Chapter MA, Problem 1.8MA

To determine

To express: The unit vector $a_n$ makes angles $\alpha ,\beta \text{ and }\gamma$ with the coordinate x-, y-, and z-axes, respectively. Express a non-unit vector $\overrightarrow{OP}$ of length $\mathcal{l}$ that is parallel to $a_n$.

Answer to Problem 1.8MA

The unit vector $a_n$ makes angles $\alpha ,\beta \text{ and }\gamma$ with the coordinate x-, y-, and z-axes, respectively is expressed as $\underset{\_}{\cos \alpha a_x+\cos \beta a_y+\cos \gamma a_z}$ and a non-unit vector $\overrightarrow{OP}$ of length $\mathcal{l}$ that is parallel to $a_n$ is expressed as $\underset{\_}{l\cos \alpha a_x+l\cos \beta a_y+l\cos \gamma a_z}$.

Explanation of Solution

Suppose the unit vector is $a_n=x^{\prime }{a}_x+y^{\prime }{a}_y+z^{\prime }{a}_z$.

Perform dot product with $a_x$ on both sides of $a_n=x^{\prime }{a}_x+y^{\prime }{a}_y+z^{\prime }{a}_z$.

$\begin{array}{l}{a}_n\cdot a_x=\left[x^{\prime }{a}_x+y^{\prime }{a}_y+z^{\prime }{a}_z\right]\cdot a_x\\ \left|a_n\right|\left|a_x\right|\cos \alpha =x^{\prime }\left[\begin{array}{l}\because a_x\cdot a_x=1,a_x\cdot a_y=0\\ a_x\cdot a_z=0\end{array}\right]\\ x^{\prime }=\cos \alpha \left[\because a_n\text{ and }{a}_x\text{ are unit vectors}\right]\end{array}$

Perform dot product with $a_y$ on both sides of $a_n=x^{\prime }{a}_x+y^{\prime }{a}_y+z^{\prime }{a}_z$.

$\begin{array}{l}{a}_n\cdot a_y=\left[x^{\prime }{a}_x+y^{\prime }{a}_y+z^{\prime }{a}_z\right]\cdot a_y\\ \left|a_n\right|\left|a_y\right|\cos \beta =y^{\prime }\left[\begin{array}{l}\because a_x\cdot a_x=1,a_x\cdot a_y=0\\ a_x\cdot a_z=0\end{array}\right]\\ y^{\prime }=\cos \beta \left[\because a_n\text{ and }{a}_y\text{ are unit vectors}\right]\end{array}$

Perform dot product with $a_z$ on both sides of $a_n=x^{\prime }{a}_x+y^{\prime }{a}_y+z^{\prime }{a}_z$.

$\begin{array}{l}{a}_n\cdot a_z=\left[x^{\prime }{a}_x+y^{\prime }{a}_y+z^{\prime }{a}_z\right]\cdot a_z\\ \left|a_n\right|\left|a_z\right|\cos \gamma =z^{\prime }\left[\begin{array}{l}\because a_x\cdot a_x=1,a_x\cdot a_y=0\\ a_x\cdot a_z=0\end{array}\right]\\ z^{\prime }=\cos \gamma \left[\because a_n\text{ and }{a}_z\text{ are unit vectors}\right]\end{array}$

Thus, the unit vector $a_n$ can be expressed as $a_n=\cos \alpha a_x+\cos \beta a_y+\cos \gamma a_z$.

Consider the non-unit vector $\overrightarrow{OP}$ as $\overrightarrow{OP}=p{a}_x+q{a}_y+r{a}_z$.

Perform dot product with $a_n$ on both sides of $\overrightarrow{OP}=p{a}_x+q{a}_y+r{a}_z$.

$\begin{array}{l}\overrightarrow{OP}\cdot a_n=\left[p{a}_x+q{a}_y+r{a}_z\right]\cdot a_n\\ \left|\overrightarrow{OP}\right|\left|a_n\right|\cos 0°=\left[p{a}_x+q{a}_y+r{a}_z\right]\cdot \left[\cos \alpha a_x+\cos \beta a_y+\cos \gamma a_z\right]\\ \mathcal{l}=p\cos \alpha +q\cos \beta +r\cos \gamma \left[\begin{array}{l}\because a_x\cdot a_x=1,a_x\cdot a_y=0,a_x\cdot a_z=0,\\ a_n\text{ is a unit vector and }\left|\overrightarrow{OP}\right|=\mathcal{l}\end{array}\right]\\ \end{array}$

Perform cross product with $a_n$ on both sides of $\overrightarrow{OP}=p{a}_x+q{a}_y+r{a}_z$.

$\begin{array}{l}\overrightarrow{OP}\times a_n=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ p& q& r\\ \cos \alpha & \cos \beta & \cos \gamma \end{array}\right|\\ \left|\overrightarrow{OP}\right|\left|a_n\right|\sin 0°a_n=a_x\left[q\cos \gamma -r\cos \beta \right]-a_y\left[p\cos \gamma -r\cos \alpha \right]\\ +a_z\left[p\cos \beta -q\cos \alpha \right]\left[\because a_n\text{ is a unit vector and }\left|\overrightarrow{OP}\right|=\mathcal{l}\right]\\ a_x\left[q\cos \gamma -r\cos \beta \right]-a_y\left[p\cos \gamma -r\cos \alpha \right]+a_z\left[p\cos \beta -q\cos \alpha \right]=0\\ \end{array}$

Then, the system of equations is obtained as

$\begin{array}{l}q\cos \gamma -r\cos \beta =0\mathrm{...}\left(1\right)\\ p\cos \gamma -r\cos \alpha =0\mathrm{...}\left(2\right)\\ p\cos \beta -q\cos \alpha =0\mathrm{...}\left(3\right)\end{array}$

From the equation 1 and 2, obtain the values of r and equate them.

$\begin{array}{l}r=\frac{q\cos \gamma }{\cos \beta }\text{ and }r=\frac{p\cos \gamma }{\cos \alpha }\\ \frac{p\cos \gamma }{\cos \alpha }=\frac{q\cos \gamma }{\cos \beta }\left(\text{Divide both sides by cos}\gamma \right)\\ p=\frac{q\cos \alpha }{\cos \beta }\end{array}$

Substitute the value of p in equation 2 and obtain the value of r.

$\begin{array}{l}\left(\frac{q\cos \alpha }{\cos \beta }\right)\cos \gamma -r\cos \alpha =0\left(\because p=\frac{q\cos \alpha }{\cos \beta }\right)\\ r\cos \alpha =\frac{q\cos \alpha \cos \gamma }{\cos \beta }\left(\text{Divide both sides by }\cos \alpha \right)\\ r=\frac{q\cos \gamma }{\cos \beta }\end{array}$

Substitute the values of p and r in the equation $\mathcal{l}=p\cos \alpha +q\cos \beta +r\cos \gamma$ and obtain the value of q as follows.

$\begin{array}{l}\mathcal{l}=\left(\frac{q\cos \alpha }{\cos \beta }\right)\cos \alpha +q\cos \beta +\left(\frac{q\cos \gamma }{\cos \beta }\right)\cos \gamma \\ \mathcal{l}\cos \beta =q{\cos }^2\alpha +q{\cos }^2\beta +q{\cos }^2\gamma \\ q\left({\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma \right)=\mathcal{l}\cos \beta \\ q=\mathcal{l}\cos \beta \left(\because \left|a_n\right|=\sqrt{{\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma }=1\right)\end{array}$

Substitute back the value of q in p and r in order to obtain their values.

That is,

$\begin{array}{l}p=\frac{l\cos \beta \cos \alpha }{\cos \beta }=l\cos \alpha \\ r=\frac{l\cos \beta \cos \gamma }{\cos \beta }=l\cos \gamma \end{array}$

Thus, the non-unit vector $\overrightarrow{OP}$ as $\overrightarrow{OP}=p{a}_x+q{a}_y+r{a}_z$ becomes $\overrightarrow{OP}=l\cos \alpha a_x+l\cos \beta a_y+l\cos \gamma a_z$.

Therefore, the unit vector $a_n$ makes angles $\alpha ,\beta \text{ and }\gamma$ with the coordinate x-, y-, and z-axes, respectively is expressed as $\underset{\_}{\cos \alpha a_x+\cos \beta a_y+\cos \gamma a_z}$ and a non-unit vector $\overrightarrow{OP}$ of length $\mathcal{l}$ that is parallel to $a_n$ is expressed as $\underset{\_}{l\cos \alpha a_x+l\cos \beta a_y+l\cos \gamma a_z}$.