Question

If $\left(1,-2,-2\right)$ and $\left(0,2,1\right)$ are direction ratios of two lines, then the direction cosines of a perpendicular to both the lines are

• A. $\left(\frac{1}{3},-\frac{1}{3},\frac{2}{3}\right)$
• B. $\left(\frac{2}{3},-\frac{1}{3},\frac{2}{3}\right)$
• C. $\left(-\frac{2}{3},-\frac{1}{3},\frac{2}{3}\right)$
• D. $\left(\frac{2}{\sqrt{14}},-\frac{1}{\sqrt{14}},\frac{3}{\sqrt{14}}\right)$

Answer 1

The correct option is B

$\left(\frac{2}{3},-\frac{1}{3},\frac{2}{3}\right)$

Explanation for correct option:

Step-1: Finding the vector which is perpendicular on two given lines.

Given, direction ratio of the lines are $\left(1,-2,-2\right)$ and $\left(0,2,1\right)$.

We know that a Vector which is perpendicular to the lines which have direction ratio are $\left(a_1,b_1,c_1\right)\&\left(a_2,b_2,c_2\right)$ is $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|$

Vector Which is perpendicular to the lines whose direction ratio are $\left(1,-2,-2\right)$ and $\left(0,2,1\right)$.is $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& -2\\ 0& 2& 1\end{array}\right|$

$=2\hat{i}-\hat{j}+2\hat{k}$

Step-2: Finding direction cosine.

We know that the direction cosine of a vector $a\hat{i}+b\hat{j}+c\hat{k}$ is $\Rightarrow \left(\frac{a}{\sqrt{a^2+b^2+c^2}},\frac{b}{\sqrt{a^2+b^2+c^2}},\frac{c}{\sqrt{a^2+b^2+c^2}}\right)$

Therefore required direction ratio is $\Rightarrow \left(\frac{2}{\sqrt{{\left(2\right)}^2+{\left(-1\right)}^2+{\left(2\right)}^2}},\frac{-1}{\sqrt{{\left(2\right)}^2+{\left(-1\right)}^2+{\left(2\right)}^2}},\frac{2}{\sqrt{{\left(2\right)}^2+{\left(-1\right)}^2+{\left(2\right)}^2}}\right)$

$\Rightarrow \left(\frac{2}{3},\frac{-1}{3},\frac{2}{3}\right)$

Hence, correct answer is option $\left(B\right)$