Question

The direction ratios of lines intersecting the line $\dfrac{{x - 3}}{2} = \dfrac{{y - 3}}{1} = \dfrac{z}{1}$ at angle ${60^\circ}$ are:
A. $1,2, - 1$
B. $1, - 1,2$
C. $1,1,2$
D. $1, - 2,1$

Answer 1

To solve this question, first we will assume a Cartesian vector of two lines. Then we will apply the formula to find the angle between both the lines. And then we will get an equation. Use options to satisfy the equation.

Complete step by step solution:
Let the direction ratios be $x,y\,and\,z$ . And the vector is $xi + yj + zk$ .
And the coefficients of the given direction ratio are 2, 1 and 1.
Now, the angle between the two lines is:
$\cos \theta = \dfrac{{\mathop {{b_1}}\limits^ \to .\mathop {{b_2}}\limits^ \to }}{{|\mathop {{b_1}}\limits^ \to |.|\mathop {{b_2}}\limits^ \to |}}$
$\Rightarrow \cos {60^\circ} = \dfrac{{2x + y + z}}{{\sqrt {{2^2} + {2^2} + {1^2}} \times \sqrt {{x^2} + {y^2} + {z^2}} }}$
$\Rightarrow \dfrac{1}{2} = \dfrac{{2x + y + z}}{{\sqrt 6 \sqrt {{x^2} + {y^2} + {z^2}} }}$
$\Rightarrow \dfrac{1}{2} = \dfrac{{2x + y + z}}{{\sqrt 6 .\sqrt {{x^2} + {y^2} + {z^2}} }}$
Now, satisfy the options one-by-one:
In option A.: $1,2, - 1$. Put the direction ratios in the upper equation:
$\therefore \dfrac{{2 \times 1 + 1 \times 2 + ( - 1)}}{{\sqrt 6 .\sqrt {{1^2} + {2^2} + {{( - 1)}^2}} }} = \dfrac{3}{{\sqrt 6 .\sqrt 6 }} = \dfrac{3}{6} = \dfrac{1}{2}$ is equals to $\dfrac{1}{2}$ .
So, option A. is satisfying the equation that we found.
Hence, the correct option is (A.) $1,2, - 1$ .

Note:
Now that we have understood what direction cosines are, we can move to direction ratios. Any numbers that are proportional to the direction cosines are called direction ratios, usually represented as $x,\,y\,and\,z$ .