Question

The equation of a line is $5x-3=15y+7=3-10z$ . Find the direction cosines of the given line.

Answer 1

Hint: The direction ratios of the line \[\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}\] is a, b, c. hence we will write the given equation in this form and first find direction ratios. Now if a, b, c. are direction ratios of line then direction cosines are given by $\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$ , $\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$ , $\dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$

Complete step by step answer:
Now Direction cosines of the line are basically the cosine angles of the line made with coordinate axes. Hence if α, β and γ are the angles of the line made by coordinate axis x, y, z respectively then the direction cosines of the line is $\cos \alpha ,\cos \beta ,\cos \gamma $
To find direction cosines we first need to find direction ratios of the line.
Direction ratios of the line \[\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}\] is a, b, c. hence we will try to write the equation in this form.
Now Consider the equation of given line $5x-3=15y+7=3-10z$
We can also write this equation as
$5\left( x-\dfrac{3}{5} \right)=15\left( y+\dfrac{7}{15} \right)=-10\left( z-\dfrac{3}{10} \right)$
Now we will take LCM of 5, 15, 10 which is 30. Hence dividing the whole equation by 30 we get
$\dfrac{\left( x-\dfrac{3}{5} \right)}{6}=\dfrac{\left( y+\dfrac{7}{15} \right)}{2}=\dfrac{\left( z-\dfrac{3}{10} \right)}{-3}$
Now the above equation is in the form of \[\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}\]
Hence we get the direction ratios as 6, 2, -3.
Now if a, b, c. are direction ratios of line then direction cosines are given by $\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$ , $\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$ , $\dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$
Hence the direction cosine of the given lines become $\dfrac{6}{\sqrt{{{6}^{2}}+{{2}^{2}}+{{(-3)}^{2}}}}$, $\dfrac{2}{\sqrt{{{6}^{2}}+{{2}^{2}}+{{(-3)}^{2}}}}$ , $\dfrac{-3}{\sqrt{{{6}^{2}}+{{2}^{2}}+{{(-3)}^{2}}}}$
Hence solving this we get the direction cosines as
$=\dfrac{6}{\sqrt{36+4+9}}$, $\dfrac{2}{\sqrt{36+4+9}}$ , $\dfrac{-3}{\sqrt{36+4+9}}$
$=\dfrac{6}{\sqrt{49}}$ , $\dfrac{2}{\sqrt{49}}$ , $\dfrac{-3}{\sqrt{49}}$
$=\dfrac{6}{7}$ , $\dfrac{2}{7}$ , $\dfrac{-3}{7}$

Hence the direction cosines of the lines are $=\dfrac{6}{7}$ , $\dfrac{2}{7}$ , $\dfrac{-3}{7}$

Note: Direction ratios of the line \[\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}\] is a, b, c which should not be confused with direction cosines which are $\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$, $\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$ , $\dfrac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$.