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Three Dimensional Geometry - Direction cosines and direction ratios of a line

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Direction Angles: A directed line LL passing through the origin makes angles α\alpha, β\beta, and γ\gamma with the positive directions of the X,Y,X, Y, and ZZ axes respectively. These are known as direction angles. Visually, if you imagine a ray starting from the origin, α\alpha is the angle it tilts away from the horizontal X-axis, β\beta from the vertical Y-axis, and γ\gamma from the depth-oriented Z-axis.

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Direction Cosines (DCs): The cosine values of the direction angles, denoted as l=cos⁡α,m=cos⁡β,l = \cos \alpha, m = \cos \beta, and n=cos⁡γn = \cos \gamma, are the direction cosines of the line. A key geometric property is that for any line, the sum of the squares of its direction cosines is always unity: l2+m2+n2=1l^2 + m^2 + n^2 = 1. This implies that the vector (l,m,n)(l, m, n) is a unit vector.

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Direction Ratios (DRs): Any three numbers a,b,ca, b, c that are proportional to the direction cosines l,m,nl, m, n are called direction ratios. Unlike DCs, which are restricted by the unit sum property, a line can have infinitely many sets of DRs. Visually, DRs can be viewed as the components of any vector that lies parallel to the line, representing the relative change in x,y,x, y, and zz coordinates.

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Relationship between DCs and DRs: If a,b,ca, b, c are the direction ratios of a line, the direction cosines are calculated by dividing each ratio by the magnitude a2+b2+c2\sqrt{a^2 + b^2 + c^2}. This process 'normalizes' the ratios. Depending on the direction of the line, the DCs are given by l=±aa2+b2+c2l = \pm \frac{a}{\sqrt{a^2 + b^2 + c^2}}, m=±ba2+b2+c2m = \pm \frac{b}{\sqrt{a^2 + b^2 + c^2}}, and n=±ca2+b2+c2n = \pm \frac{c}{\sqrt{a^2 + b^2 + c^2}}.

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Direction Cosines of a Line Joining Two Points: For a line segment passing through points P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2), the direction ratios are proportional to the differences in coordinates: (x2−x1),(y2−y1),(x_2 - x_1), (y_2 - y_1), and (z2−z1)(z_2 - z_1). The distance PQ=(x2−x1)2+(y2−y1)2+(z2−z1)2PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} serves as the normalizing factor to find the DCs.

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Direction Cosines of Coordinate Axes: For the X-axis, the direction angles are 0∘,90∘,90∘0^{\circ}, 90^{\circ}, 90^{\circ}, resulting in DCs of (1,0,0)(1, 0, 0). Similarly, the Y-axis has DCs of (0,1,0)(0, 1, 0) and the Z-axis has DCs of (0,0,1)(0, 0, 1). This visually represents that each axis has no component or 'tilt' toward the other two perpendicular axes.

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Sign and Orientation: A line in 3D space extends infinitely in two opposite directions. Therefore, a line has two sets of direction cosines. If (l,m,n)(l, m, n) represents the direction cosines for one direction, then (−l,−m,−n)(-l, -m, -n) represents the direction cosines for the opposite direction.

📐Formulae

l=cos⁡α,m=cos⁡β,n=cos⁡γl = \cos \alpha, m = \cos \beta, n = \cos \gamma

l2+m2+n2=1l^2 + m^2 + n^2 = 1

la=mb=nc\frac{l}{a} = \frac{m}{b} = \frac{n}{c}

l=±aa2+b2+c2,m=±ba2+b2+c2,n=±ca2+b2+c2l = \pm \frac{a}{\sqrt{a^2 + b^2 + c^2}}, m = \pm \frac{b}{\sqrt{a^2 + b^2 + c^2}}, n = \pm \frac{c}{\sqrt{a^2 + b^2 + c^2}}

PQ=(x2−x1)2+(y2−y1)2+(z2−z1)2PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

l=x2−x1PQ,m=y2−y1PQ,n=z2−z1PQl = \frac{x_2 - x_1}{PQ}, m = \frac{y_2 - y_1}{PQ}, n = \frac{z_2 - z_1}{PQ}

💡Examples

Problem 1:

Find the direction cosines of a line that makes equal angles with the three positive coordinate axes.

Solution:

Let the angles made by the line with the X,Y,X, Y, and ZZ axes be α,β,\alpha, \beta, and γ\gamma. Since the angles are equal, α=β=γ\alpha = \beta = \gamma. Thus, l=cos⁡α,m=cos⁡α,n=cos⁡αl = \cos \alpha, m = \cos \alpha, n = \cos \alpha. \n Using the identity l2+m2+n2=1l^2 + m^2 + n^2 = 1: \n cos⁡2α+cos⁡2α+cos⁡2α=1\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1 \n 3cos⁡2α=13 \cos^2 \alpha = 1 \n cos⁡2α=13\cos^2 \alpha = \frac{1}{3} \n cos⁡α=±13\cos \alpha = \pm \frac{1}{\sqrt{3}} \n Therefore, the direction cosines are (±13,±13,±13)(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}).

Explanation:

Because the angles are identical, the direction cosines must be identical. We solve for the common value using the fundamental identity that the sum of squares of DCs is 1.

Problem 2:

If a line has direction ratios 2,−1,−22, -1, -2, determine its direction cosines.

Solution:

Given the direction ratios a=2,b=−1,c=−2a = 2, b = -1, c = -2. \n Step 1: Calculate the magnitude a2+b2+c2\sqrt{a^2 + b^2 + c^2}. \n 22+(−1)2+(−2)2=4+1+4=9=3\sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3. \n Step 2: Calculate l,m,nl, m, n using the ratios. \n l=aa2+b2+c2=23l = \frac{a}{\sqrt{a^2+b^2+c^2}} = \frac{2}{3} \n m=ba2+b2+c2=−13m = \frac{b}{\sqrt{a^2+b^2+c^2}} = \frac{-1}{3} \n n=ca2+b2+c2=−23n = \frac{c}{\sqrt{a^2+b^2+c^2}} = \frac{-2}{3} \n The direction cosines are (23,−13,−23)(\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}).

Explanation:

Direction cosines are found by dividing each direction ratio by the square root of the sum of the squares of the ratios, which normalizes the vector to unit length.