CBSE Board Examination 2024 · Section A · 1 mark

Q9: Find $k$ when the direction cosines are $\sqrt{3}k, \sqrt{3}k, \sqrt{3}k$

The question

If the direction cosines of a line are $\sqrt{3}\,k$ ⁠, $\sqrt{3}\,k$ ⁠, $\sqrt{3}\,k$ ⁠, then the value of k is :

(A) $\pm 1$

(B) $\pm\sqrt{3}$

(D) $\pm\frac{1}{3}$

Exam answer

For direction cosines $l, m, n$ ⁠: $l^2 + m^2 + n^2 = 1$ ⁠.

Why?

The direction cosines $l, m, n$ are the components of a unit vector along the line, and a unit vector has length $1$ ⁠. So

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

Watch out: this holds for direction cosines only, not for direction ratios.

Here $l = m = n = \sqrt{3}\,k$ ⁠, so

\begin{aligned} 3k^2 + 3k^2 + 3k^2 &= 1 \\ \Rightarrow 9k^2 &= 1 \\ \Rightarrow k^2 &= \frac{1}{9} \\ \Rightarrow k &= \pm\frac{1}{3} \end{aligned}

Why?

Each of $l^2$ ⁠, $m^2$ ⁠, $n^2$ is $(\sqrt{3}\,k)^2 = 3k^2$ ⁠, because $(\sqrt{3})^2 = 3$ ⁠.

\begin{aligned} (\sqrt{3}\,k)^2 &= (\sqrt{3})^2 \times k^2 \\ &= 3k^2 \\[4pt] &3k^2 + 3k^2 + 3k^2 \\ &= (3 + 3 + 3)k^2 = 9k^2 \\[4pt] 9k^2 &= 1 \\ k^2 &= \frac19 \quad (\text{divide by } 9) \\ k &= \pm\sqrt{\frac19} \\ &= \pm\frac{\sqrt1}{\sqrt9} = \pm\frac13 \end{aligned}

A positive number has two square roots, so $k = 1/3$ or $k = -1/3$ ⁠. Both are valid: they give the same line, described in opposite directions.

Answer: (D) $\pm\dfrac{1}{3}$

Where this comes from

Chapter 11, Three Dimensional Geometry: Direction cosines of a line.

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Three Dimensional Geometry in past papers: question 6 of 35 · see them all

Q9: Find kkk when the direction cosines are 3k,3k,3k\sqrt{3}k, \sqrt{3}k, \sqrt{3}k3​k,3​k,3​k