Question

The projections of a line on the co-ordinate axes are \[4,{\text{ }}6,{\text{ }}12\]. The direction cosines of the line are
A. $\dfrac{2}{7},\dfrac{3}{7},\dfrac{6}{7}$
B. $2,3,6$
C. $\dfrac{2}{{11}},\dfrac{3}{{11}},\dfrac{6}{{11}}$
D. None of these

Answer 1

Hint: Given, the projections of a line on the co-ordinate axes are \[4,{\text{ }}6,{\text{ }}12\]. We have to find the direction cosine of the line. First, we will assume the position vector of a given line segment. Then, we will find the direction cosine of the given line using formula $\left( {\dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right)$.

Formula used: If a vector is given by $\vec {A} = a\hat i + b\hat j + c\hat k$, then
 direction cosine = $\left( {\dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right)$

Complete step by step solution:
Given, the projections of a line on the co-ordinate axes are \[4,{\text{ }}6,{\text{ }}12\].
Let us assume the position vector of the given line segment endpoints are $\vec P$ and $\vec Q$
Now, we have given the projections of a line on the co-ordinate axes are \[4,{\text{ }}6,{\text{ }}12\].
So, $\vec {PQ} = 4\hat i + 6\hat j + 12\hat k$
The direction cosines (also known as directional cosines) of a vector in analytical geometry are the cosines of the angles the vector makes with the three positive coordinate axes. They are, in essence, the contributions of each basis component to a unit vector pointing in that direction.
Now, we will find the direction cosine of the given line segment
Direction cosines are $\dfrac{4}{{\sqrt {{4^2} + {6^2} + {{12}^2}} }},\dfrac{6}{{\sqrt {{4^2} + {6^2} + {{12}^2}} }},\dfrac{{12}}{{\sqrt {{4^2} + {6^2} + {{12}^2}} }}$
On solving the above expression
$\dfrac{4}{{\sqrt {16 + 36 + 144} }},\dfrac{6}{{\sqrt {16 + 36 + 144} }},\dfrac{{12}}{{\sqrt {16 + 36 + 144} }}$
On further solving
$\dfrac{4}{{\sqrt {196} }},\dfrac{6}{{\sqrt {196} }},\dfrac{{12}}{{\sqrt {196} }}$
We know $\sqrt {196} = 14$
$ \Rightarrow \dfrac{4}{{14}},\dfrac{6}{{14}},\dfrac{{12}}{{14}}$
After simplifying
$\dfrac{2}{7},\dfrac{3}{7},\dfrac{6}{7}$
Hence, direction cosines of given line are $\dfrac{2}{7},\dfrac{3}{7},\dfrac{6}{7}$

So, option (A) is the correct answer.

Note: Students can make calculation mistakes while solving the question so they should calculate that correctly. And apply direction cosine formula $\left( {\dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right)$ correctly to get accurate answer. Should solve the solution thoroughly, do not skip any step to get an exact solution.