Question

If a line has direction ratios 2, -1, -2, then what are its direction cosines?

Answer 1

We have given the value of direction ratios as 2, -1, -2 and we are asked to find the direction cosines. Let us assume that direction ratios are a, b and c and the direction cosines are l, m and n. Then the relationship between direction ratios and direction cosines are: $l=\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}},m=\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}},n=\dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$. Now, substituting the direction ratios in this formula will give you the value of direction cosines.

Complete step by step answer:
We have given the direction ratios as 2, -1, -2 and we are asked to find the direction cosines.
Let us assume that direction ratios are a, b and c and direction cosines as l, m and n.
There is a relationship between direction ratios and direction cosines as follows:
$\begin{aligned} & l=\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}, \\\ & m=\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}, \\\ & n=\dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \\\ \end{aligned}$
Now, we have given the direction ratios in the above problem as 2, -1 and -2. Let us consider the value of a, b and c as 2, -1 and -2 respectively and substituting these values in the above equations we get,
$\begin{aligned} & l=\dfrac{2}{\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( -2 \right)}^{2}}}} \\\ & \Rightarrow l=\dfrac{2}{\sqrt{4+1+4}} \\\ & \Rightarrow l=\dfrac{2}{\sqrt{9}} \\\ \end{aligned}$
We know that the square root of 9 is 3.
$l=\dfrac{2}{3}$
From the above solution, we have got the value of one of the direction cosines as $\dfrac{2}{3}$.
$\begin{aligned} & m=\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \\\ & \Rightarrow m=\dfrac{-1}{\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( -2 \right)}^{2}}}} \\\ & \Rightarrow m=\dfrac{-1}{\sqrt{4+1+4}} \\\ & \Rightarrow m=\dfrac{-1}{\sqrt{9}} \\\ & \Rightarrow m=-\dfrac{1}{3} \\\ \end{aligned}$
From the above solution, we have got the value of one of the direction cosines as $-\dfrac{1}{3}$.
$\begin{aligned} & n=\dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \\\ & \Rightarrow n=\dfrac{-2}{\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( -2 \right)}^{2}}}} \\\ & \Rightarrow n=\dfrac{-2}{\sqrt{4+1+4}} \\\ & \Rightarrow n=\dfrac{-2}{\sqrt{9}} \\\ & \Rightarrow n=-\dfrac{2}{3} \\\ \end{aligned}$

Hence, we have got the value of direction cosines as $\dfrac{2}{3},-\dfrac{1}{3},-\dfrac{2}{3}$.

Note: We can cross check the values of direction cosines that we got above are correct or not by using the following relationship.
We have shown above the relation between direction ratios and direction cosines as:
$\begin{aligned} & l=\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}, \\\ & m=\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}, \\\ & n=\dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}} \\\ \end{aligned}$
If we take square on both the sides of the above equations and add them we get,
$\begin{aligned} & {{l}^{2}}=\dfrac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}, \\\ & {{m}^{2}}=\dfrac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}, \\\ & {{n}^{2}}=\dfrac{{{c}^{2}}}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \\\ \end{aligned}$
Now, adding the above three equations we get,

& {{l}^{2}}+{{m}^{2}}+{{n}^{2}}=\dfrac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}+\dfrac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}+\dfrac{{{c}^{2}}}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}} \\\ & \Rightarrow {{l}^{2}}+{{m}^{2}}+{{n}^{2}}=\dfrac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}=1 \\\ \end{aligned}$$ The relation that we got from the above calculations is: $${{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1$$ Now, substituting the values of direction cosines that we have solved above $l=\dfrac{2}{3},m=-\dfrac{1}{3},n=-\dfrac{2}{3}$ in the above equation and check whether these values of l, m and n are satisfying the above relation or not. $$\begin{aligned} & {{\left( \dfrac{2}{3} \right)}^{2}}+{{\left( -\dfrac{1}{3} \right)}^{2}}+{{\left( -\dfrac{2}{3} \right)}^{2}}=1 \\\ & \Rightarrow \dfrac{4}{9}+\dfrac{1}{9}+\dfrac{4}{9}=1 \\\ & \Rightarrow \dfrac{4+1+4}{9}=1 \\\ & \Rightarrow \dfrac{9}{9}=1 \\\ & \Rightarrow 1=1 \\\ \end{aligned}$$ As you can see that L.H.S is equal to R.H.S so the values of l, m and n that we have calculated is correct.