CBSE Sample Paper 2023-24 · Section A · 1 mark

Q18: $c$ when the direction cosines are $\frac{1}{c}, \frac{1}{c}, \frac{1}{c}$

Sample 2023-24Ch 11 · Three Dimensional Geometry

The question

If the direction cosines of a line are $\left\langle \frac{1}{c}, \frac{1}{c}, \frac{1}{c} \right\rangle$ ⁠, then

(a) $0 < c < 1$

(b) $c > 2$

(d) $c = \pm\sqrt 3$

Exam answer

\begin{aligned} l^2 + m^2 + n^2 &= 1 \\ \Rightarrow \frac{1}{c^2} + \frac{1}{c^2} + \frac{1}{c^2} &= 1 \\ \Rightarrow \frac{3}{c^2} &= 1 \\ \Rightarrow c^2 &= 3 \end{aligned}

Why?

If a line makes angles $\alpha, \beta, \gamma$ with the axes, its direction cosines are $l = \cos\alpha$ ⁠, $m = \cos\beta$ ⁠, $n = \cos\gamma$ ⁠, and always

l^2 + m^2 + n^2 = 1

Here $l = m = n = \frac{1}{c}$ ⁠, so the three squares add up to $\frac{3}{c^2}$ ⁠.

\begin{aligned} \left(\frac{1}{c}\right)^2 &= \frac{1^2}{c^2} = \frac{1}{c^2} \\ &\frac{1}{c^2} + \frac{1}{c^2} + \frac{1}{c^2} \\ &= \frac{1 + 1 + 1}{c^2} = \frac{3}{c^2} \\[4pt] \frac{3}{c^2} &= 1 \\ 3 &= c^2 \quad (\text{multiply by } c^2) \end{aligned}

Take the square root of both sides of $c^2 = 3$ ⁠, with both signs:

c = \pm\sqrt 3

$\therefore\ c = \pm\sqrt 3$

Why?

Both signs are allowed: reversing the direction of the line changes the signs of all three direction cosines.

Check: $\left(\frac{1}{\sqrt 3}\right)^2 \times 3 = \frac{1}{3} \times 3 = 1$ ⁠. Options (a), (b) and (c) give $\frac{3}{c^2} \ne 1$ ⁠.

Answer: (d) $c = \pm\sqrt 3$

Where this comes from

Chapter 11, Three Dimensional Geometry: Direction cosines: l² + m² + n² = 1.

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Three Dimensional Geometry in past papers: question 24 of 35 · see them all

Q18: ccc when the direction cosines are 1c,1c,1c\frac{1}{c}, \frac{1}{c}, \frac{1}{c}c1​,c1​,c1​

Q18: $c$ when the direction cosines are $\frac{1}{c}, \frac{1}{c}, \frac{1}{c}$