Question

If the direction cosines of a line are $\dfrac{1}{c}$ , $\dfrac{1}{c}$ , $\dfrac{1}{c}$ , then
A. $0 < c < 1$
B. $c > 2$
C. $c > 0$
D. $c = \pm \sqrt 3 $

Answer 1

Hint: We have to use the relation between the direction cosine of a given line to get the value of $c$ . In a 3D space, we can compute the direction cosines for a given vector or a straight line. Direction cosines are represented by $l$, $m$, $n$ with respect to the angle that the line has made with the $x$, $y$, $z$-axis.

Formula used: Direction cosines can be determined using formula:
$l = \cos \alpha $ , $m = \cos \beta $ and $n = \cos \gamma $ .

Complete step-by-step solution:
We have direction cosines of a line $\dfrac{1}{c}$ , $\dfrac{1}{c}$ , $\dfrac{1}{c}$
Consider a point $(x,y,z)$ at a distance of $r$ from the origin, hence the distance between two points will be
${r^2} = {(x - 0)^2} + {(y - 0)^2} + {(z - 0)^2}$, solving this we get
${r^2} = {x^2} + {y^2} + {z^2}$
And we know that the direction cosines are represented as
$l = \cos \alpha = \dfrac{x}{r}$ , $m = \cos \beta = \dfrac{y}{r}$ and $n = \cos \gamma = \dfrac{z}{r}$ .
Adding the square of these three values, we get
${l^2} + {m^2} + {n^2} = {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = \dfrac{{{x^2}}}{{{r^2}}} + \dfrac{{{y^2}}}{{{r^2}}} + \dfrac{{{z^2}}}{{{r^2}}}$
Solving this we get
${l^2} + {m^2} + {n^2} = 1$ ……..(1.1)
So, we have $(l,m,n) = \left( {\dfrac{1}{c},\dfrac{1}{c},\dfrac{1}{c}} \right)$
Substituting this value in equation (1.1)
$\dfrac{1}{{{c^2}}} + \dfrac{1}{{{c^2}}} + \dfrac{1}{{{c^2}}} = 1$
$\dfrac{3}{{{c^2}}} = 1$
On solving further, we get
$c = \pm \sqrt 3 $
Hence, the correct option is D.

Note: We must comprehend that the positive axis of each of the three coordinates is used to determine each angle. Any vector’s relationship to any of the coordinate axes can be determined using these direction cosines. In a similar manner we can determine the $x$-axis’ direction cosine which is $(l,m,n) = (1,0,0)$ . Similarly, for $y$-axis $(l,m,n) = (0,1,0)$ and for $z$-axis $(l,m,n) = (0,0,1)$.