Question

If the length of a vector is $21$ and the direction ratios are $2, - 3,6$ , then its direction cosines are:
A. $\dfrac{2}{{21}},\dfrac{{ - 1}}{7},\dfrac{2}{7}$
B. $\dfrac{2}{7},\dfrac{{ - 3}}{7},\dfrac{6}{7}$
C. $\dfrac{2}{7},\dfrac{3}{7},\dfrac{6}{7}$
D. None of these

Answer 1

Hint: In order to solve this type of question, first we will write the direction ratios given to us. Next, we will write the direction cosines of a vector. Then, we will substitute the values of direction ratios in the formula of direction cosines. Next, we will simplify it further to get the correct answer.

Formula used:
$\cos \alpha = \dfrac{x}{{\left| {\sqrt {{x^2} + {y^2} + {z^2}} } \right|}}$
$\cos \beta = \dfrac{y}{{\left| {\sqrt {{x^2} + {y^2} + {z^2}} } \right|}}$
$\cos \gamma = \dfrac{z}{{\left| {\sqrt {{x^2} + {y^2} + {z^2}} } \right|}}$

Complete step by step solution:
We are given that,
$x = 2,$ $y = - 3,$ $z = 6$ ………………..equation $\left( 1 \right)$
We know that direction cosines are given by,
$\cos \alpha = \dfrac{x}{{\left| {\sqrt {{x^2} + {y^2} + {z^2}} } \right|}}$
$\cos \beta = \dfrac{y}{{\left| {\sqrt {{x^2} + {y^2} + {z^2}} } \right|}}$
$\cos \gamma = \dfrac{z}{{\left| {\sqrt {{x^2} + {y^2} + {z^2}} } \right|}}$
Substituting the values from equation $\left( 1 \right)$ to above equations
\[\cos \alpha = \dfrac{2}{{\left| {\sqrt {{2^2} + {{\left( { - 3} \right)}^2} + {6^2}} } \right|}}\]
$\cos \alpha = \dfrac{2}{7}$
Solving for $\cos \beta ,$
\[\cos \beta = \dfrac{{ - 3}}{{\left| {\sqrt {{2^2} + {{\left( { - 3} \right)}^2} + {6^2}} } \right|}}\]
$\cos \beta = \dfrac{{ - 3}}{7}$
Solving for $\cos \gamma ,$
\[\cos \gamma = \dfrac{6}{{\left| {\sqrt {{2^2} + {{\left( { - 3} \right)}^2} + {6^2}} } \right|}}\]
$\cos \gamma = \dfrac{6}{7}$
Thus, the direction cosines are $\dfrac{2}{7},\dfrac{{ - 3}}{7},\dfrac{6}{7}$.
$\therefore $ The correct option is (B).

Note: When a line passes through origin in three dimensions with x-axis, y-axis and z-axis making angles \[\alpha\],\[\beta \] and $\gamma $ with it then it is termed as direction angles. And when we take cosines of these angles then it is termed as direction cosines. To solve these we must remember the formulas and its concept.