maths > vector-algebra

Directional Cosine

what you'll learn...

Overview

Directional Cosine

»  Directional cosines:

→  $\stackrel{\to }{p}=ai+bj+ck$$\vec{p} = a i + b j + c k$ makes angles $\alpha ,\beta ,\gamma$$\alpha , \beta , \gamma$ with $x,y,z$$x , y , z$-axes respectively

→  $\mathrm{cos}\alpha =\frac{a}{|p|}$$\cos \alpha = \frac{a}{| p |}$

→  $\mathrm{cos}\beta =\frac{b}{|p|}$$\cos \beta = \frac{b}{| p |}$

→  $\mathrm{cos}\gamma =\frac{c}{|p|}$$\cos \gamma = \frac{c}{| p |}$

alternate representation

'cosine ' of an angle is 'the ratio of adjacent side (adjacent to the given angle) to the hypotenuse in a right angled triangle'.

While referring to vector quantities, we use the form that specifies the components along the axes. eg: $ai+bj+ck$$a i + b j + c k$ specifies the component along $x-$$x -$axis is $a$$a$, the component along $y-$$y -$axis is $b$$b$, and the component along $z-$$z -$axis is $c$$c$.

The alternate representation to component form is to specify magnitude and angles made by the vector on the axes. This has the necessary information about the vector.

To describe the vector $ai+bj$$a i + b j$ shown in the figure, if '$r$$r$ and $\theta$$\theta$' are available we can derive the vector form $ai+bj$$a i + b j$.

It is noted that the values of $a$$a$ and $b$$b$ in the figure are '$a=8\mathrm{cos}60$$a = 8 \cos 60$ and $b=8\mathrm{sin}60$$b = 8 \sin 60$'.

The same in 3D, to describe the vector $ai+bj+ck$$a i + b j + c k$ shown in figure, the vector is completely defined by '$r$$r$ and $\alpha ,\beta$$\alpha , \beta$'. If these three parameters are available we can derive the vector notation $ai+bj+ck$$a i + b j + c k$.

Note that the third angle can be derived from the other two angles.

Given the vector in figure, values of $a$$a$, $b$$b$, and $c$$c$ are derived as $a=3\mathrm{cos}60$$a = 3 \cos 60$; $b=3\mathrm{cos}45$$b = 3 \cos 45$; $c=3\mathrm{cos}75$$c = 3 \cos 75$

When a vector with magnitude $r$$r$ and angles $\alpha ,\beta ,\gamma$$\alpha , \beta , \gamma$ is given, the coordinate form of the vector is
$r\left(\mathrm{cos}\alpha i+\mathrm{cos}\beta j+\mathrm{cos}\gamma k\right)$$r \left(\cos \alpha i + \cos \beta j + \cos \gamma k\right)$
$\mathrm{cos}\alpha$$\cos \alpha$, $\mathrm{cos}\beta$$\cos \beta$, $\mathrm{cos}\gamma$$\cos \gamma$ are called the directional cosines of the vector.

Directional cosines: Given that a vector $\stackrel{\to }{p}=ai+bj+ck$$\vec{p} = a i + b j + c k$ makes angles $\alpha ,\beta ,\gamma$$\alpha , \beta , \gamma$ with $x,y,z$$x , y , z$-axes respectively, then the directional cosines of the vector are
$\mathrm{cos}\alpha =\frac{a}{|p|}$$\cos \alpha = \frac{a}{| p |}$
$\mathrm{cos}\beta =\frac{b}{|p|}$$\cos \beta = \frac{b}{| p |}$
$\mathrm{cos}\gamma =\frac{c}{|p|}$$\cos \gamma = \frac{c}{| p |}$

Directional cosines are the ratio of "projections on to an axes" to "the magnitude".

The representation of vectors in the component form is $\stackrel{\to }{p}=ai+bj+ck$$\vec{p} = a i + b j + c k$
Directional cosines along with magnitude provide an alternate representation of a vector.
$\stackrel{\to }{p}=r\left(\mathrm{cos}\alpha i+\mathrm{cos}\beta j+\mathrm{cos}\gamma k\right)$$\vec{p} = r \left(\cos \alpha i + \cos \beta j + \cos \gamma k\right)$

$\mathrm{cos}\alpha =\frac{a}{r}$$\cos \alpha = \frac{a}{r}$

$\mathrm{cos}\beta =\frac{b}{r}$$\cos \beta = \frac{b}{r}$

$\mathrm{cos}\gamma =\frac{c}{r}$$\cos \gamma = \frac{c}{r}$

${r}^{2}={a}^{2}+{b}^{2}+{c}^{2}$${r}^{2} = {a}^{2} + {b}^{2} + {c}^{2}$

${\mathrm{cos}}^{2}\alpha +{\mathrm{cos}}^{2}\beta +{\mathrm{cos}}^{2}\gamma =1$${\cos}^{2} \alpha + {\cos}^{2} \beta + {\cos}^{2} \gamma = 1$

Directional cosines make Unit Vector: For a given vector $ai+bj+ck$$a i + b j + c k$ the directional cosine vector $li+mj+nk$$l i + m j + n k$ is the unit vector in the direction of the given vector. Note that

$l=\mathrm{cos}\alpha =\frac{a}{r}$$l = \cos \alpha = \frac{a}{r}$

$m=\mathrm{cos}\beta =\frac{b}{r}$$m = \cos \beta = \frac{b}{r}$

$n=\mathrm{cos}\gamma =\frac{c}{r}$$n = \cos \gamma = \frac{c}{r}$
This also implies that

${l}^{2}+{m}^{2}+{n}^{2}=1$${l}^{2} + {m}^{2} + {n}^{2} = 1$

examples

What is the magnitude of the directional cosines vector : $\mathrm{cos}\alpha i+\mathrm{cos}\beta j+\mathrm{cos}\gamma k$$\cos \alpha i + \cos \beta j + \cos \gamma k$ ?

The answer is '$1$$1$'. The magnitude = $\sqrt{{\mathrm{cos}}^{2}\alpha +{\mathrm{cos}}^{2}\beta +{\mathrm{cos}}^{2}\gamma }$$\sqrt{{\cos}^{2} \alpha + {\cos}^{2} \beta + {\cos}^{2} \gamma}$, which evaluates to 1.

Find the direction cosines of a line which makes equal angles with the coordinate axes.

$\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$$\left(\frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}}\right)$ OR $\left(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$$\left(- \frac{1}{\sqrt{3}} , - \frac{1}{\sqrt{3}} , - \frac{1}{\sqrt{3}}\right)$

From the question, we know that $\alpha =\beta =\gamma$$\alpha = \beta = \gamma$ and We know the property of directional cosines ${\mathrm{cos}}^{2}\alpha +{\mathrm{cos}}^{2}\beta +{\mathrm{cos}}^{2}\gamma =1$${\cos}^{2} \alpha + {\cos}^{2} \beta + {\cos}^{2} \gamma = 1$

If a line makes angles $45,135,90$$45 , 135 , 90$ with the x, y and z-axes respectively, find its direction cosines.

The answer is '$\mathrm{cos}45,\mathrm{cos}135,\mathrm{cos}90$$\cos 45 , \cos 135 , \cos 90$'.

Find the directional cosine of the vector $2i-3j+\sqrt{3}k$$2 i - 3 j + \sqrt{3} k$.

magnitude of the vector = 4
So directional cosines are $\frac{2}{4},-\frac{3}{4},\frac{\sqrt{3}}{4}$$\frac{2}{4} , - \frac{3}{4} , \frac{\sqrt{3}}{4}$

summary

Directional cosines: Given that a vector $\stackrel{\to }{p}=ai+bj+ck$$\vec{p} = a i + b j + c k$ makes angles $\alpha ,\beta ,\gamma$$\alpha , \beta , \gamma$ with $x,y,z$$x , y , z$-axes respectively, then the directional cosines of the vector are
$\mathrm{cos}\alpha =\frac{a}{|p|}$$\cos \alpha = \frac{a}{| p |}$
$\mathrm{cos}\beta =\frac{b}{|p|}$$\cos \beta = \frac{b}{| p |}$
$\mathrm{cos}\gamma =\frac{c}{|p|}$$\cos \gamma = \frac{c}{| p |}$

Directional cosines are the ratio of "projections on to an axes" to "the magnitude".

Directional cosines make Unit Vector: For a given vector $ai+bj+ck$$a i + b j + c k$ the directional cosine vector $li+mj+nk$$l i + m j + n k$ is the unit vector in the direction of the given vector. Note that

$l=\mathrm{cos}\alpha =\frac{a}{r}$$l = \cos \alpha = \frac{a}{r}$

$m=\mathrm{cos}\beta =\frac{b}{r}$$m = \cos \beta = \frac{b}{r}$

$n=\mathrm{cos}\gamma =\frac{c}{r}$$n = \cos \gamma = \frac{c}{r}$
This also implies that

${l}^{2}+{m}^{2}+{n}^{2}=1$${l}^{2} + {m}^{2} + {n}^{2} = 1$

Outline