Question

The projection of a vector on the three coordinate axes are (6, -3, 2) respectively. The direction cosines of the vector are:
${\text{A}}{\text{. 6, - 3, 2}}$
${\text{B}}{\text{. }}\dfrac{6}{5},{\text{ }}\dfrac{{ - 3}}{5},\dfrac{2}{5}$
${\text{C}}{\text{. }}\dfrac{6}{7},\dfrac{{ - 3}}{7},\dfrac{2}{7}$
${\text{D}}{\text{. }}\dfrac{{ - 6}}{7},\dfrac{{ - 3}}{7},\dfrac{2}{7}$

Answer 1

**** Go with the definition of the direction cosines. Take the under root sum of squares of the 3 projections and then divide the individual projection by that.

Complete step-by-step answer :

Direction cosines denote the cosines of the angles between the vector and the three coordinate axes. They are the contributions of each component (projection) of the basis to a unit vector in that direction.

So, if we have a vector $\vec a$ with its components along the three axes as ${a_x}\hat i,{a_y}\hat j,{a_z}\hat k$, then the direction cosines namely, l, m, n of the vector $\vec a$ would be-

$l = \cos \alpha = \dfrac{{{a_x}}}{{\sqrt {{a_x}^2 + {a_y}^2 + {a_z}^2} }}$

$m = \cos \beta = \dfrac{{{a_y}}}{{\sqrt {{a_x}^2 + {a_y}^2 + {a_z}^2} }}$

$n = \cos \gamma = \dfrac{{{a_z}}}{{\sqrt {{a_x}^2 + {a_y}^2 + {a_z}^2} }}$

So, following the above concept-

For the given vector the projections are- (6, -3, 2)

So, the direction cosines would be-

$l = \dfrac{{ 6}}{{\sqrt {{{\left( 6 \right)}^2} + {{\left( { - 3} \right)}^2} + {{\left( 2 \right)}^2}} }} = \dfrac{ 6}{7}$

$m = \dfrac{{ - 3}}{{\sqrt {{{\left( 6 \right)}^2} + {{\left( { - 3} \right)}^2} + {{\left( 2 \right)}^2}} }} = \dfrac{ - 3}{7}$

$n = \dfrac{{ 2}}{{\sqrt {{{\left( 6 \right)}^2} + {{\left( { - 3} \right)}^2} + {{\left( 2 \right)}^2}} }} = \dfrac{ 2}{7}$

Therefore, the correct option is C.

Note**:** Direction ratios are proportional to the direction cosines and are represented by the letters . Direction cosines are concerned with the unit vector while direction ratios are with their multiples.