Direction Cosines and Direction Ratios of a Line

Introduction to 3D Geometry and Direction Angles

In three-dimensional geometry, describing the orientation of a straight line is one of the first essential ideas. Consider a directed line $L$. If this line makes angles $\alpha$, $\beta$, and $\gamma$ with the positive directions of the x-, y-, and z-axes respectively, then these are called the direction angles of the directed line.

• If a line does not pass through the origin, we can always draw a line through the origin parallel to it. Since parallel lines have the same orientation, they have the same direction angles.
• If the direction of the line is reversed, then the new direction angles become $\pi-\alpha$, $\pi-\beta$, and $\pi-\gamma$.
• Thus, a line has two opposite orientations, and changing the orientation changes the signs of its direction cosines.

This idea is important because in 3D geometry, we often study a line through its direction rather than its exact position.

Direction Cosines of a Line

If $\alpha$, $\beta$, and $\gamma$ are the direction angles of a directed line, then the numbers $$\cos\alpha, \quad \cos\beta, \quad \cos\gamma$$ are called the direction cosines of the line. These are usually denoted by $l, m, n$ respectively: $$l = \cos\alpha, \qquad m = \cos\beta, \qquad n = \cos\gamma.$$

These three numbers completely determine the direction of the line.

Fundamental Identity: The direction cosines always satisfy $$l^2 + m^2 + n^2 = 1.$$

Why this is true: Take any point $P(x,y,z)$ on the line at distance $r$ from the origin. Then the coordinates of $P$ can be written as $$x = lr, \qquad y = mr, \qquad z = nr.$$ Using the distance formula, $$r^2 = x^2 + y^2 + z^2.$$ Substituting $x=lr, y=mr, z=nr$ gives $$r^2 = l^2r^2 + m^2r^2 + n^2r^2 = r^2(l^2+m^2+n^2).$$ Since $r \neq 0$ for any non-origin point on the line, dividing by $r^2$ yields $$l^2+m^2+n^2=1.$$

This identity is the most important test for checking whether a given triple can be direction cosines.

Direction Ratios of a Line

Any three numbers $a, b, c$ which are proportional to the direction cosines $l, m, n$ of a line are called the direction ratios of the line. Thus, for some non-zero constant $k$, $$l = ak, \qquad m = bk, \qquad n = ck.$$ Equivalently, $$\frac{l}{a} = \frac{m}{b} = \frac{n}{c} = k.$$

Unlike direction cosines, direction ratios do not need to satisfy any special identity like $a^2+b^2+c^2=1$. Any three real numbers, not all zero, can serve as direction ratios.

Converting Direction Ratios to Direction Cosines

Since $$l=ak,\quad m=bk,\quad n=ck,$$ and also $$l^2+m^2+n^2=1,$$ we get $$a^2k^2+b^2k^2+c^2k^2=1$$ $$k^2(a^2+b^2+c^2)=1.$$ So, $$k = \pm \frac{1}{\sqrt{a^2+b^2+c^2}}.$$ Hence the direction cosines are $$l = \pm \frac{a}{\sqrt{a^2+b^2+c^2}}, \qquad m = \pm \frac{b}{\sqrt{a^2+b^2+c^2}}, \qquad n = \pm \frac{c}{\sqrt{a^2+b^2+c^2}}.$$

The same sign must be taken throughout for one orientation of the line. Taking the opposite common sign gives the opposite orientation.

Direction Cosines and Ratios of a Line Joining Two Points

Let a line pass through two points $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$. Then the vector $$\overrightarrow{PQ} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$$ gives the direction of the line from $P$ to $Q$.

Therefore, the direction ratios of the line segment $PQ$ are $$a = x_2 - x_1, \qquad b = y_2 - y_1, \qquad c = z_2 - z_1.$$

If the length of $PQ$ is $$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2},$$ then the direction cosines of the directed line from $P$ to $Q$ are $$l = \frac{x_2-x_1}{d}, \qquad m = \frac{y_2-y_1}{d}, \qquad n = \frac{z_2-z_1}{d}.$$

If we reverse the direction and consider $Q$ to $P$, then each of these changes sign.

Collinearity Test using Direction Ratios

Three points $A, B, C$ are collinear if the direction ratios of $AB$ are proportional to the direction ratios of $BC$ (or $AC$). In other words, the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ must be parallel.

Example 1: Finding Direction Cosines from Angles

If a line makes angles $90^\circ, 60^\circ$, and $30^\circ$ with the positive directions of x-, y-, and z-axes respectively, find its direction cosines.

Solution: Step 1: Write the given direction angles: $$\alpha = 90^\circ, \qquad \beta = 60^\circ, \qquad \gamma = 30^\circ.$$ Step 2: Use the definition $$l = \cos\alpha, \qquad m = \cos\beta, \qquad n = \cos\gamma.$$ Step 3: Evaluate each cosine: $$l = \cos 90^\circ = 0,$$ $$m = \cos 60^\circ = \frac{1}{2},$$ $$n = \cos 30^\circ = \frac{\sqrt{3}}{2}.$$ Step 4: Verify the identity: $$l^2+m^2+n^2 = 0^2 + \left(\frac12\right)^2 + \left(\frac{\sqrt3}{2}\right)^2 = 0 + \frac14 + \frac34 = 1.$$ So the values are correct.

Answer: $0, \frac{1}{2}, \frac{\sqrt{3}}{2}$

Example 2: Converting Direction Ratios to Direction Cosines

If a line has direction ratios $2, -1, -2$, determine its direction cosines.

Solution: Step 1: Let the direction ratios be $$a=2, \qquad b=-1, \qquad c=-2.$$ Step 2: Compute $$\sqrt{a^2+b^2+c^2} = \sqrt{2^2+(-1)^2+(-2)^2} = \sqrt{4+1+4} = \sqrt9 = 3.$$ Step 3: Divide each direction ratio by 3 to get one set of direction cosines: $$l = \frac{2}{3}, \qquad m = -\frac{1}{3}, \qquad n = -\frac{2}{3}.$$ Step 4: Since a line can be taken in the opposite direction as well, the opposite set is also valid: $$-\frac23, \qquad \frac13, \qquad \frac23.$$ If the direction is not specified, either set represents the same line.

Answer: One possible set is $\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}$; the opposite set is also valid.

Example 3: Line Joining Two Points

Find the direction cosines of the line passing through the two points $(-2, 4, -5)$ and $(1, 2, 3)$.

Solution: Step 1: Let $$P(-2,4,-5), \qquad Q(1,2,3).$$ Then the direction ratios of the directed line from $P$ to $Q$ are $$a = 1-(-2)=3, \qquad b = 2-4=-2, \qquad c = 3-(-5)=8.$$ So the direction ratios are $3,-2,8$.

Step 2: Find the length of $PQ$: $$PQ = \sqrt{3^2+(-2)^2+8^2} = \sqrt{9+4+64} = \sqrt{77}.$$ Step 3: Divide each direction ratio by $\sqrt{77}$: $$l = \frac{3}{\sqrt{77}}, \qquad m = -\frac{2}{\sqrt{77}}, \qquad n = \frac{8}{\sqrt{77}}.$$ Step 4: Check: $$\left(\frac{3}{\sqrt{77}}\right)^2 + \left(-\frac{2}{\sqrt{77}}\right)^2 + \left(\frac{8}{\sqrt{77}}\right)^2 = \frac{9+4+64}{77} = 1.$$ Hence these are correct.

Answer: For the directed line from $P$ to $Q$, the direction cosines are $\frac{3}{\sqrt{77}}, -\frac{2}{\sqrt{77}}, \frac{8}{\sqrt{77}}$.

Example 4: Checking Collinearity

Show that the points $A(2, 3, -4)$, $B(1, -2, 3)$, and $C(3, 8, -11)$ are collinear.

Solution: Step 1: Find the direction ratios of $AB$: $$\overrightarrow{AB} = (1-2,\,-2-3,\,3-(-4)) = (-1,-5,7).$$ So DRs of $AB$ are $-1,-5,7$.

Step 2: Find the direction ratios of $BC$: $$\overrightarrow{BC} = (3-1,\,8-(-2),\,-11-3) = (2,10,-14).$$ So DRs of $BC$ are $2,10,-14$.

Step 3: Check proportionality: $$\frac{-1}{2} = -\frac12, \qquad \frac{-5}{10} = -\frac12, \qquad \frac{7}{-14} = -\frac12.$$ All three ratios are equal.

Step 4: Therefore, $$\overrightarrow{AB} \parallel \overrightarrow{BC}.$$ Since both segments share the common point $B$, the three points must lie on the same straight line.

Answer: Proved. The points are collinear.

Example 5: Finding an Unknown Direction Angle

If a line makes angles $45^\circ$ and $60^\circ$ with the positive directions of x- and y-axes respectively, find the angle it makes with the positive z-axis.

Solution: Step 1: Given $$\alpha = 45^\circ, \qquad \beta = 60^\circ.$$ We need to find $\gamma$.

Step 2: Use the identity $$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1.$$ Substitute the values: $$\cos^2 45^\circ + \cos^2 60^\circ + \cos^2\gamma = 1.$$ That is, $$\left(\frac{1}{\sqrt2}\right)^2 + \left(\frac12\right)^2 + \cos^2\gamma = 1$$ $$\frac12 + \frac14 + \cos^2\gamma = 1$$ $$\frac34 + \cos^2\gamma = 1$$ $$\cos^2\gamma = \frac14.$$ Step 3: Hence $$\cos\gamma = \pm \frac12.$$ Therefore, $$\gamma = 60^\circ \quad \text{or} \quad 120^\circ.$$ Both are possible because the line may be directed in two opposite orientations with respect to the z-axis.

Answer: $60^\circ$ or $120^\circ$

Example 6: Trigonometric Identity with Direction Angles

Show that if a line makes angles $\alpha, \beta, \gamma$ with the coordinate axes, then $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2$.

Solution: Step 1: Start with the standard identity for direction cosines: $$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1.$$ Step 2: Use $$\cos^2\theta = 1 - \sin^2\theta.$$ Then $$(1-\sin^2\alpha) + (1-\sin^2\beta) + (1-\sin^2\gamma) = 1.$$ Step 3: Simplify: $$3 - (\sin^2\alpha + \sin^2\beta + \sin^2\gamma) = 1.$$ Step 4: Rearranging, $$\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2.$$ Thus the required relation is proved.

Answer: Proved.

Example 7: Line Equally Inclined to Axes

Find the direction cosines of a line which is equally inclined to the positive coordinate axes.

Solution: Step 1: If the line is equally inclined to the positive x-, y-, and z-axes, then the direction angles are equal, so the direction cosines are equal: $$l = m = n.$$ Step 2: Use the identity $$l^2 + m^2 + n^2 = 1.$$ Since $l=m=n$, $$3l^2 = 1 \implies l^2 = \frac13 \implies l = \frac{1}{\sqrt3}.$$ We take the positive value because the line is equally inclined to the positive axes.

Step 3: Therefore, $$l = m = n = \frac{1}{\sqrt3}.$$

Answer: $\left(\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, \frac{1}{\sqrt3}\right)$

Example 8: Finding Point Coordinates from DCs and Distance

A line passes through the origin and has direction cosines $\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}$. Find the coordinates of the point on this line which is at a distance of 6 units from the origin.

Solution: Step 1: Let the required point be $P(x,y,z)$. Since the line passes through the origin and the distance of the point from the origin is $r=6$, we use $$x = lr, \qquad y = mr, \qquad z = nr.$$ Step 2: Substitute the given direction cosines: $$x = \frac13 \cdot 6 = 2,$$ $$y = -\frac23 \cdot 6 = -4,$$ $$z = \frac23 \cdot 6 = 4.$$ So one point is $$P(2,-4,4).$$ Step 3: Since the line extends in the opposite direction also, the point at the same distance on the opposite side of the origin is $$(-2,4,-4).$$

Answer: $(2,-4,4)$ or $(-2,4,-4)$

Example 9: Direction Cosines of the Coordinate Axes

What are the direction cosines of the x-, y-, and z-axes?

Solution: Step 1: For the x-axis, the direction angles are $$0^\circ, 90^\circ, 90^\circ.$$ Therefore, $$\cos 0^\circ, \cos 90^\circ, \cos 90^\circ = 1,0,0.$$ So the direction cosines of the x-axis are $(1,0,0)$.

Step 2: For the y-axis, the direction angles are $$90^\circ, 0^\circ, 90^\circ.$$ Therefore, the direction cosines are $(0,1,0)$.

Step 3: For the z-axis, the direction angles are $$90^\circ, 90^\circ, 0^\circ.$$ Therefore, the direction cosines are $(0,0,1)$.

Answer: x-axis: $(1,0,0)$; y-axis: $(0,1,0)$; z-axis: $(0,0,1)$

Example 10: Direction Ratios of a Perpendicular Line

Find the direction ratios of a line perpendicular to two lines whose direction ratios are $1, 2, 3$ and $-2, 1, 4$.

Solution: Step 1: A vector perpendicular to both given lines must be perpendicular to the vectors $$\vec{u} = \langle 1,2,3 \rangle, \qquad \vec{v} = \langle -2,1,4 \rangle.$$ So its direction ratios can be obtained from the cross product $\vec{u} \times \vec{v}$.

Step 2: Compute the cross product: $$\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -2 & 1 & 4 \end{vmatrix}$$ $$= \hat{i}(2\cdot 4 - 3\cdot 1) - \hat{j}(1\cdot 4 - 3\cdot(-2)) + \hat{k}(1\cdot 1 - 2\cdot(-2))$$ $$= \hat{i}(8-3) - \hat{j}(4+6) + \hat{k}(1+4)$$ $$= 5\hat{i} - 10\hat{j} + 5\hat{k}.$$ Step 3: Therefore, one set of direction ratios is $$5, -10, 5,$$ which simplifies to $$1, -2, 1.$$ The opposite set $-1,2,-1$ is also valid.

Answer: One possible set of direction ratios is $1,-2,1$.