Question

The direction cosines of a line equally inclined to three mutually perpendicular lines having direction cosines as ${l_1},{m_1},{n_1}$ , ${l_2},{m_2},{n_2}$ and ${l_3},{m_3},{n_3}$ are:
A. \[{l_1} + {l_2} + {l_3}\] , \[{m_1} + {m_2} + {m_3}\] , \[{n_1} + {n_2} + {n_3}\]
B. $\dfrac{{{l_1} + {l_2} + {l_3}}}{{\sqrt 3 }}$ , $\dfrac{{{m_1} + {m_2} + {m_3}}}{{\sqrt 3 }}$ , $\dfrac{{{n_1} + {n_2} + {n_3}}}{{\sqrt 3 }}$
C. $\dfrac{{{l_1} + {l_2} + {l_3}}}{3}$ , $\dfrac{{{m_1} + {m_2} + {m_3}}}{3}$ , $\dfrac{{{n_1} + {n_2} + {n_3}}}{3}$
D. None of these.

Answer 1

Hint: If we have a vector $\overrightarrow r $ or a line $ax + by + c = 0$ which makes angles $\alpha $ , $\beta $ and $\gamma $ with the $x$-axis, $y$-axis and $z$-axis respectively . Then $\cos \alpha $ , $\cos \beta $ , $\cos \gamma $ are called the direction cosines of the given vector or line. They are usually denoted by $l$ , $m$ , $n$ such that $l = \cos \alpha $ , $m = \cos \beta $ and $n = \cos \gamma $ .

Formula Used: The relation between direction cosines is given as ${l^2} + {m^2} + {n^2} = 1$ .

Complete step-by-step solution:
The direction cosines of three lines that are mutually perpendicular to one another are known as ${l_1},{m_1},{n_1}$ , ${l_2},{m_2},{n_2}$ and ${l_3},{m_3},{n_3}$ , respectively. Therefore,
${l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2} = 0$
${l_2}{l_3} + {m_2}{m_3} + {n_2}{n_3} = 0$
And
${l_1}{l_3} + {m_1}{m_3} + {n_1}{n_3} = 0$
i.e., $\sqrt {{l_1}^2 + {m_1}^2 + {n_1}^2} = 1$ ,
$\sqrt {{l_2}^2 + {m_2}^2 + {n_2}^2} = 1$ and similarly,
$\sqrt {{l_3}^2 + {m_3}^2 + {n_3}^2} = 1$ .
So, we can express the sum of these three equations together as:
$
  {({l_1} + {l_2} + {l_3})^2} + {({m_1} + {m_2} + {m_3})^2} + {({n_1} + {n_2} + {n_3})^2} \\
   = ({l_1}^2 + {m_1}^2 + {n_1}^2) + ({l_2}^2 + {m_2}^2 + {n_2}^2) + ({l_3}^2 + {m_3}^2 + {n_3}^2) + 2({l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2}) + 2({l_2}{l_3} + {m_2}{m_3} + {n_2}{n_3}) + 2({l_1}{l_3} + {m_1}{m_3} + {n_1}{n_3}) \\
 $
Substituting the values from above equations, we get
${({l_1} + {l_2} + {l_3})^2} + {({m_1} + {m_2} + {m_3})^2} + {({n_1} + {n_2} + {n_3})^2} = 3$
The above equation can also be written as
${({l_1} + {l_2} + {l_3})^2} + {({m_1} + {m_2} + {m_3})^2} + {({n_1} + {n_2} + {n_3})^2} = {(\sqrt 3 )^2}$
We know that ${l^2} + {m^2} + {n^2} = 1$ hence, to get the value of direction cosines we divide the whole equation by ${(\sqrt 3 )^2}$ on both sides, then we have the direction cosines as follows:
$\dfrac{{{l_1} + {l_2} + {l_3}}}{{\sqrt 3 }}$ , $\dfrac{{{m_1} + {m_2} + {m_3}}}{{\sqrt 3 }}$ , $\dfrac{{{n_1} + {n_2} + {n_3}}}{{\sqrt 3 }}$

Hence, the correct option is B.

Note: Let $\overrightarrow r = (x,y,z)$ be a position vector having direction cosines as $\cos \alpha $ , $\cos \beta $ , $\cos \gamma $ then, the direction cosines will be represented as follows:
$l = \cos \alpha = \dfrac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }}$ , $m = \cos \beta = \dfrac{y}{{\sqrt {{x^2} + {y^2} + {z^2}} }}$ and $n = \cos \gamma = \dfrac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }}$ .