Question

If $\left( {\dfrac{1}{2},\dfrac{1}{3},n} \right)$ are the direction cosines of a line then what is the value of $n$ ?
A. $\dfrac{{\sqrt {23} }}{6}$
B. $\dfrac{{23}}{6}$
C. $\dfrac{2}{3}$
D. $\dfrac{3}{2}$

Answer 1

Hint: For vectors in space, their direction is determined by the angles made by the vector with respect to the x-axis, y-axis, and z-axis. The cosine of these angles results in the direction cosines of that vector. Let $l$, $m$, and $n$ be the direction cosines of a vector, then the sum of the squares of these direction cosines equals 1. Therefore, ${l^2} + {m^2} + {n^2} = 1$ .

Complete step by step Solution:
As we know that the sum of the squares of the direction cosines of a line equals 1, that is for a line with direction cosines as $(l,m,n)$, then
${l^2} + {m^2} + {n^2} = 1$
Therefore, for the given direction cosines $\left( {\dfrac{1}{2},\dfrac{1}{3},n} \right)$ of the line,
${\left( {\dfrac{1}{2}} \right)^2} + {\left( {\dfrac{1}{3}} \right)^2} + {n^2} = 1$
Keeping the variables on the left side of the above equation,
${n^2} = 1 - \dfrac{1}{4} - \dfrac{1}{9}$
Simplifying further,
${n^2} = \dfrac{{36 - 13}}{{36}}$
Taking the square root,
$n = \pm \dfrac{{\sqrt {23} }}{6}$
As only $\dfrac{{\sqrt {23} }}{6}$ is given in the options,
Hence, $n = \dfrac{{\sqrt {23} }}{6}$

Hence, the correct option is (A).

Note: The direction cosines of a line need not be only positive. Direction cosines of a vector are the cosines of the angles the coplanar vector made with the x-axis, y-axis, and z-axis respectively. They can be both positive and negative.