Question

The direction ratios of two lines are 1, -2,-2 and 0, 2, 1. The direction cosines of the line perpendicular to the above lines are
A. $\dfrac{2}{3}, - \dfrac{1}{3},\dfrac{2}{3}$
B. $\dfrac{{ - 1}}{3},\dfrac{2}{3},\dfrac{2}{3}$
C. $\dfrac{1}{4},\dfrac{3}{4},\dfrac{1}{2}$
D.None of these

Answer 1

Hint : We need to find the direction cosine of the line perpendicular to the two lines, whose direction ratios are given. We use the fact that if direction ratios $({a_1},{b_1},{c_1})$ is perpendicular to direction ratios of $({a_2},{b_2},{c_2})$ , then ${a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3} = 0$ . We use Cramer’s rule to solve linear equations.

Complete step-by-step answer :
Let the direction ratios of lines perpendicular to both the lines be $(a,b,c)$ .
The direction ratios of two lines are given: $(1, - 2, - 2)$ and $(0,2,1)$ .
We know $({a_1},{b_1},{c_1}) \bot ({a_2},{b_2},{c_2})$ $\Rightarrow {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3} = 0$
Since $(a,b,c) \bot (1, - 2, - 2)$ and $(a,b,c) \bot (0,2,1)$ . That is both lines are perpendicular to $(a,b,c)$ .
Now take, $(a,b,c) \bot (1, - 2, - 2)$ then we get:
$\Rightarrow (a \times 1) + (b \times - 2) + (c \times - 2) = 0$
$\Rightarrow a - 2b - 2c = 0$ ---- (1)
Now take, $(a,b,c) \bot (0,2,1)$ then we get:
$\Rightarrow (a \times 0) + (b \times 2) + (c \times 1) = 0$
$\Rightarrow 0.a + 2b + c = 0$ ----- (2)
We solve equation (1) and (2) by Cramer’s rule.
That is, if we have two linear equations ${a_1}a + {b_1}b + {c_1}c = 0$ and ${a_2}a + {b_2}b + {c_2}c = 0$ then
$\Rightarrow \dfrac{a}{{({b_1}{c_2} - {c_1}{b_2})}} = \dfrac{b}{{({c_1}{a_2} - {a_1}{c_2})}} = \dfrac{c}{{({a_1}{b_2} - {b_1}{a_2})}}$
Now applying the same rule for $a - 2b - 2c = 0$ and $0.a + 2b + c = 0$ .
Comparing with above we have ${a_1} = 1$ , ${b_1} = - 2$ , ${c_1} = - 2$ , ${a_2} = 0$ , ${b_2} = 2$ and ${c_2} = 1$ .
$\Rightarrow \dfrac{a}{{\left( {( - 2 \times 1) - ( - 2 \times 2)} \right)}} = \dfrac{b}{{\left( {( - 2 \times 0) - (1 \times 1)} \right)}} = \dfrac{c}{{\left( {(1 \times 2) - ( - 2 \times 0)} \right)}}$
$\Rightarrow \dfrac{a}{{ - 2 + 4}} = \dfrac{b}{{0 - 1}} = \dfrac{c}{{2 - 0}}$
$\Rightarrow \dfrac{a}{2} = \dfrac{b}{{ - 1}} = \dfrac{c}{2}$
Hence the ratio is $(a,b,c) = (2, - 1,2)$ .
Now to find the direction cosine: If $(a,b,c)$ are direction ratios and direction cosine are given by,
$\Rightarrow \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$
$\Rightarrow \dfrac{2}{{\sqrt {{2^2} + {{( - 1)}^2} + {2^2}} }},\dfrac{{ - 1}}{{\sqrt {{2^2} + {{( - 1)}^2} + {2^2}} }},\dfrac{2}{{\sqrt {{2^2} + {{( - 1)}^2} + {2^2}} }}$
$\Rightarrow \dfrac{2}{{\sqrt {4 + 1 + 4} }},\dfrac{{ - 1}}{{\sqrt {4 + 1 + 4} }},\dfrac{2}{{\sqrt {4 + 1 + 4} }}$
$\Rightarrow \dfrac{2}{{\sqrt 9 }},\dfrac{{ - 1}}{{\sqrt 9 }},\dfrac{2}{{\sqrt 9 }}$
$\Rightarrow \dfrac{2}{3}, - \dfrac{1}{3},\dfrac{2}{3}$
So, the correct answer is “Option A”.

Note : Remember how to find direction cosine if they are given ration ratios. Know the difference between direction ratios and direction cosine. Direction cosine is the angle made by the line with the positive coordinate axis. Direction ratios are the numbers which are proportional to the direction cosine. In above we mention the Crimson’s rule which is only for two linear equations. The formula changes for three linear equations.