Question

The direction ratios of two lines are 1, −2, −2 and 0, 2, 1, then the direction cosines of the line perpendicular to the above lines are
(a)$\left( \dfrac{1}{3},\dfrac{-1}{3},\dfrac{2}{3} \right)$
(b)$\left( \dfrac{2}{3},\dfrac{-1}{3},\dfrac{2}{3} \right)$
(c)$\left( \dfrac{-2}{3},\dfrac{-1}{3},\dfrac{2}{3} \right)$
(d)$\left( \dfrac{2}{\sqrt{14}},\dfrac{-1}{\sqrt{14}},\dfrac{3}{\sqrt{14}} \right)$

Answer 1

Hint: To find the direction cosines of the vector is needed to divide the corresponding coordinate of the vector by the length of the vector. The coordinates of the unit vector are equal to its direction cosines.

Complete step-by-step answer:
Let a, b, c be the direction ratios of the line whose direction cosines are required. Then as this line is perpendicular to the given lines so we have
a (1) + b(−2) + c(−2) = 0 and a(0) + b(2) + c(1) = 0
a - 2b - 2c = 0 and 0a + 2b + c = 0
Solving these simultaneously, we get
\[\dfrac{a}{\left| \begin{matrix}
   -2 & -2 \\
   2 & 1 \\
\end{matrix} \right|}=\dfrac{-b}{\left| \begin{matrix}
   1 & -2 \\
   0 & 1 \\
\end{matrix} \right|}=\dfrac{c}{\left| \begin{matrix}
   1 & -2 \\
   0 & 2 \\
\end{matrix} \right|}\]
$\dfrac{a}{(-2+4)}=\dfrac{-b}{(1+0)}=\dfrac{c}{(2+0)}$
$\dfrac{a}{2}=\dfrac{-b}{1}=\dfrac{c}{2}$
$\dfrac{a}{2}=\dfrac{b}{-1}=\dfrac{c}{2}$
$a:b:c=2:-1:2$
To find the direction cosines of the vector is needed to divide the corresponding coordinate of the vector by the length of the vector.
Let $l,m,n$ be the required direction cosines.
\[l=\dfrac{2}{\sqrt{4+1+4}}=\dfrac{2}{\sqrt{9}}=\dfrac{2}{3}\]
\[m=\dfrac{-1}{\sqrt{4+1+4}}=\dfrac{-1}{\sqrt{9}}=\dfrac{-1}{3}\]
\[n=\dfrac{2}{\sqrt{4+1+4}}=\dfrac{-1}{\sqrt{9}}=\dfrac{-1}{3}\]
Hence the required direction cosines are $\dfrac{2}{3},\dfrac{-1}{3},\dfrac{2}{3}$ .
Therefore, the correct option for the given question is option (b).

Note: Alternatively, the question is solved as follows
Consider a vector $\overrightarrow{a}=\widehat{i}-2\widehat{j}-2\widehat{k}$ and a vector $\overrightarrow{b}=0\widehat{i}+2\widehat{j}+\widehat{k}$. Then, the line perpendicular to the given lines will have the same direction ratios as the cross product of the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$. The cross product $\overrightarrow{a}\times \overrightarrow{b}=2\widehat{i}-\widehat{j}+2\widehat{k}$. The direction cosines of the required line will then be $\dfrac{2}{3},\dfrac{-1}{3},\dfrac{2}{3}$