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Question: If a line has direction ratios\[ - 18,{\kern 1pt} 12{\kern 1pt} ,{\kern 1pt} - 4\]then what are its ...

If a line has direction ratios−18,12,−4 - 18,{\kern 1pt} 12{\kern 1pt} ,{\kern 1pt} - 4then what are its direction cosines ?

Explanation

Solution

In this problem, we have to find direction cosines when direction ratio is given.
Let a, b, and c be the direction ratios of the given line . Then the direction cosines of the same line denoted by l, m, and n respectively is given by l=aa2+b2+c2,m=ba2+b2+c2,n=ca2+b2+c2l = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}{\kern 1pt} ,m = \dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }}{\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} n = \dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }} where l, m and n are direction cosines of the line .
This direction cosine is the angle made by the line with the x-axis , y-axis and z-axis respectively. It is proportional to the direction ratios of the line .

Complete step-by-step answer:
The given direction ratios of the line are -18 , 12, -4
So, we can use the direction cosines formulas given as follows
l=aa2+b2+c2,m=ba2+b2+c2,n=ca2+b2+c2l = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}{\kern 1pt} ,m = \dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }}{\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} n = \dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }}
i.e. let a=−18,b=12,c=−4a = - 18{\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} b = 12{\kern 1pt} ,{\kern 1pt} {\kern 1pt} c = {\kern 1pt} {\kern 1pt} - 4.
First of all, we can find the denominator

a2+b2+c2=(−18)2+122+(−4)2 a2+b2+c2=324+144+16 a2+b2+c2=484 a2+b2+c2=22  \sqrt {{a^2} + {b^2} + {c^2}} = \sqrt {{{\left( { - 18} \right)}^2} + {{12}^2} + {{\left( { - 4} \right)}^2}} \\\ \sqrt {{a^2} + {b^2} + {c^2}} = \sqrt {324 + 144 + 16} \\\ \sqrt {{a^2} + {b^2} + {c^2}} = \sqrt {484} \\\ \sqrt {{a^2} + {b^2} + {c^2}} = 22 \\\

Now we find the value of direction cosine l, m and n as follow:

l = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }} = \dfrac{{\left( { - 18} \right)}}{{22}} = \dfrac{{ - 9}}{{11}} \\\ m = \dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }} = \dfrac{{12}}{{22}} = \dfrac{6}{{11}} \\\ n = \dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }} = \dfrac{{ - 4}}{{22}} = \dfrac{{ - 2}}{{11}} \\\ $$ and Hence the direction cosines are $$\dfrac{{ - 9}}{{11}}{\kern 1pt} {\kern 1pt} ,\dfrac{6}{{11}}{\kern 1pt} ,\dfrac{{ - 2}}{{11}}$$. **Note:** Any three numbers a, b, c proportional to the direction cosines l, m, and n respectively of a line are called direction ratios of the line. Thus , we have$$\dfrac{l}{a}{\kern 1pt} {\kern 1pt} = \dfrac{m}{b}{\kern 1pt} {\kern 1pt} = \dfrac{n}{c}$$. Direction ratios of the line is proportional to direction cosines of the line . Direction cosine is the angle $$\alpha {\kern 1pt} ,{\kern 1pt} \beta {\kern 1pt} {\kern 1pt} ,\gamma $$a line makes with the x-axis , y-axis and z-axis respectively. i.e. $$l = \cos \alpha {\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} m = \cos \beta {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} {\kern 1pt} n = \cos \gamma $$. DIRECTION cosine OF x-axis are( 1,0,0) DIRECTION cosine OF y-axis are (0,1,0) DIRECTION cosine OF z-axis are (0,0,1) Summation of square of direction cosine of the line is always equal to 1.