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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 Cosines of a Line: If a line passes through the origin and makes angles α\alpha, β\beta, and γ\gamma with the positive directions of the xx, yy, and zz axes respectively, then cos⁡α\cos \alpha, cos⁡β\cos \beta, and cos⁡γ\cos \gamma are called the direction cosines of the line, denoted by ll, mm, and nn.

A 3D coordinate system showing a line L making angles alpha and beta with the X and Y axes respectively.
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Direction Ratios: Any three numbers aa, bb, and cc which are proportional to the direction cosines ll, mm, nn of a line are called its direction ratios. They satisfy the relation l=aa2+b2+c2l = \frac{a}{\sqrt{a^2+b^2+c^2}}, m=ba2+b2+c2m = \frac{b}{\sqrt{a^2+b^2+c^2}}, and n=ca2+b2+c2n = \frac{c}{\sqrt{a^2+b^2+c^2}}.

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Line Joining Two Points: For a line passing through two 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 (x2−x1)(x_2 - x_1), (y2−y1)(y_2 - y_1), and (z2−z1)(z_2 - z_1). The direction cosines are obtained by dividing these by the distance PQPQ.

A line segment connecting points P and Q in 3D space.
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Relation between DC's: The sum of the squares of the direction cosines of any line is always equal to 1, i.e., l2+m2+n2=1l^2 + m^2 + n^2 = 1.

📐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. Using the identity l2+m2+n2=1l^2 + m^2 + n^2 = 1: cos⁡2α+cos⁡2α+cos⁡2α=1\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1 3cos⁡2α=13 \cos^2 \alpha = 1 cos⁡2α=13\cos^2 \alpha = \frac{1}{3} cos⁡α=±13\cos \alpha = \pm \frac{1}{\sqrt{3}} 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. Step 1: Calculate the magnitude a2+b2+c2\sqrt{a^2 + b^2 + c^2}. 22+(−1)2+(−2)2=4+1+4=9=3\sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3. Step 2: Calculate l,m,nl, m, n using the ratios. l=aa2+b2+c2=23l = \frac{a}{\sqrt{a^2+b^2+c^2}} = \frac{2}{3} m=ba2+b2+c2=−13m = \frac{b}{\sqrt{a^2+b^2+c^2}} = \frac{-1}{3} n=ca2+b2+c2=−23n = \frac{c}{\sqrt{a^2+b^2+c^2}} = \frac{-2}{3} 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.

Problem 3:

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

Line segment AB representing the vector from A to B.

Solution:

  1. Find the direction ratios (a,b,c)(a, b, c) by calculating the differences in coordinates: a=x2−x1=1−(−2)=3a = x_2 - x_1 = 1 - (-2) = 3 b=y2−y1=2−4=−2b = y_2 - y_1 = 2 - 4 = -2 c=z2−z1=3−(−5)=8c = z_2 - z_1 = 3 - (-5) = 8
  2. Calculate the distance AB=a2+b2+c2AB = \sqrt{a^2 + b^2 + c^2}: AB=32+(−2)2+82=9+4+64=77AB = \sqrt{3^2 + (-2)^2 + 8^2} = \sqrt{9 + 4 + 64} = \sqrt{77}
  3. Calculate direction cosines (l,m,n)(l, m, n): l=377,m=−277,n=877l = \frac{3}{\sqrt{77}}, m = \frac{-2}{\sqrt{77}}, n = \frac{8}{\sqrt{77}}

Explanation:

Direction cosines of a line segment joining two points are found by dividing the differences of their respective coordinates by the distance between the two points.

Problem 4:

If a line makes angles 90∘90^{\circ}, 135∘135^{\circ}, 45∘45^{\circ} with the xx, yy and zz axes respectively, find its direction cosines.

Graph showing the orientation of the line relative to the Y-axis.

Solution:

The direction cosines are given by l=cos⁡αl = \cos \alpha, m=cos⁡βm = \cos \beta, n=cos⁡γn = \cos \gamma. Given α=90∘,β=135∘,γ=45∘\alpha = 90^{\circ}, \beta = 135^{\circ}, \gamma = 45^{\circ}. l=cos⁡90∘=0l = \cos 90^{\circ} = 0 m=cos⁡135∘=cos⁡(180∘−45∘)=−cos⁡45∘=−12m = \cos 135^{\circ} = \cos(180^{\circ} - 45^{\circ}) = -\cos 45^{\circ} = -\frac{1}{\sqrt{2}} n=cos⁡45∘=12n = \cos 45^{\circ} = \frac{1}{\sqrt{2}} The direction cosines are (0,−12,12)(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}).

Explanation:

Direct application of the definition of direction cosines using the given angles with the coordinate axes.