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Exercise 11.1 · Q3

Q.If a line has the direction ratios −18,12,−4-18, 12, -4, then what are its direction cosines ?

Punjab PsebTextbookSubjective· 2mImportance★★★★★
Appeared in past exams:CBSE 2019· Set 65/3/1· 1mexact
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Direction cosines are the cosines of the angles a line makes with the axes, found by dividing each direction ratio by the magnitude of the direction vector. For ratios −18,12,−4-18, 12, -4, the direction cosines are (−911,611,−211)\left( -\frac{9}{11}, \frac{6}{11}, -\frac{2}{11} \right).

Why Direction Ratios and Cosines?

A line in space can be described by a vector parallel to it. The components of that vector — say (a,b,c)(a, b, c) — are called direction ratios. They tell you the relative "step" the line takes along each axis, but they are not unique: scaling (a,b,c)(a, b, c) by any non-zero constant gives the same line.

Direction cosines are the unique, normalized version. They are the cosines of the angles α,β,γ\alpha, \beta, \gamma that the line makes with the xx, yy, and zz axes respectively. If a direction vector is v⃗=ai^+bj^+ck^\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}, then:

cos⁡α=aa2+b2+c2,cos⁡β=ba2+b2+c2,cos⁡γ=ca2+b2+c2\cos\alpha = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \quad \cos\beta = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \quad \cos\gamma = \frac{c}{\sqrt{a^2 + b^2 + c^2}}

The key property: cos⁡2α+cos⁡2β+cos⁡2γ=1\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1.

So the task is simple: take the given ratios, compute the magnitude, and divide each ratio by it.


Step-by-step

1. Identify the direction ratios.

We are given a=−18a = -18, b=12b = 12, c=−4c = -4.

2. Compute the magnitude (the divisor).

The magnitude of the direction vector is:

a2+b2+c2=(−18)2+122+(−4)2\sqrt{a^2 + b^2 + c^2} = \sqrt{(-18)^2 + 12^2 + (-4)^2}

=324+144+16=484=22= \sqrt{324 + 144 + 16} = \sqrt{484} = 22

Tip

484=4×121=222484 = 4 \times 121 = 22^2, so the square root is exact. Always check if the sum of squares is a perfect square — it often simplifies nicely in exam problems.

3. Divide each ratio by the magnitude. …

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