Scalars And Vectors

# Direction Cosines

Let $O$ be the origin and let $OX$, $OY$ and $OZ$ represent x-axis, y-axis and z-axis respectively. Let $P(x,y,z)$ be a point in space. Then, $\overrightarrow{OP}=(x,y,z)= x\:\overrightarrow{i}+ y\:\overrightarrow{j}+ z\:\overrightarrow{k}$ and the modulus of $\overrightarrow{OP}$ is given by, $OP=|\overrightarrow{OP}|=\sqrt{x^2+y^2+z^2}$

Let the line $\overrightarrow{OP}$ make the angles $\alpha$, $\beta$ and $\gamma$ with the three mutually perpendicular straight lines $OX$, $OY$ and $OZ$ respectively. Then, $\cos\alpha$, $\cos\beta$ and $\cos\gamma$ are said to be the direction cosines of the line $OP$. $\cos\alpha$, $\cos\beta$ and $\cos\gamma$ are denoted by $l$, $m$ and $n$ respectively.

Now, $\cos\gamma=\frac{OC}{OP}=\frac{z}{\sqrt{x^2+y^2+z^2}}$ Similarly, $\cos\alpha=\frac{x}{ \sqrt{x^2+y^2+z^2}}$ $\text{and,}\;\;\cos\beta=\frac{y}{\sqrt{x^2+y^2+z^2}}$
Also, the unit vector along $\overrightarrow{OP}$ denoted by $\hat{OP}$ is defined by, $\hat{OP}=\frac{\overrightarrow{OP}}{|\overrightarrow{OP}|}=\frac{ x\:\overrightarrow{i}+y\:\overrightarrow{j}+z\:\overrightarrow{k}}{\sqrt{x^2+y^2+z^2}}$
$=\frac{x}{\sqrt{x^2+y^2+z^2}}\overrightarrow{i}+\frac{y}{\sqrt{x^2+y^2+z^2}}\overrightarrow{j} +\frac{z}{\sqrt{x^2+y^2+z^2}}\overrightarrow{k}$
Hence, the direction cosines of the line $OP$ are the coefficients of $\overrightarrow{i}$, $\overrightarrow{j}$ and $\overrightarrow{k}$ of the unit vector $\hat{OP}$. $\therefore l=\cos\alpha =\frac{x}{\sqrt{x^2+y^2+z^2}}$ $\therefore m=\cos\beta=\frac{y}{\sqrt{x^2+y^2+z^2}}$ $\therefore n=\cos\gamma=\frac{z}{\sqrt{x^2+y^2+z^2}}$
From these values of $l$, $m$, $n$, we have, $l^2+m^2+n^2$ $=\frac{x^2}{x^2+y^2+z^2}+\frac{ y^2}{x^2+y^2+z^2}+\frac{z^2}{x^2+y^2+z^2}$ $=\frac{x^2+y^2+z^2}{x^2+y^2+z^2}=1$ $\therefore l^2+m^2+n^2=1$