Question

Find the direction cosines of the line x=4z+3, y=2-3z.

Answer 1

The direction cosines are ±$\frac{4}{\sqrt{29} }$,±$\frac{3}{\sqrt{29} }$ and ±$\frac{1}{\sqrt{29} }$

Step-by-step explanation:

The line equation is given as $x=4z+3 , y=2-3z$.

rewriting the given equation of line, $\frac{x-3}{4} =z , \frac{y-2}{-3} =z$.

Therefore, it is $\frac{x-3}{4}=\frac{y-2}{-3}=z$-----------------------------------(1)

Comparing equation 1 with $\frac{x}{a} =\frac{y}{b}=\frac{z}{c}$, where a,b,c are direction ratios.

a=4; b=-3, c=1 in equation 1,

Direction cosines are given by $\frac{a}{\sqrt{a^{2} +b^{2} +c^{2} } } , \frac{b}{\sqrt{a^{2} +b^{2} +c^{2} } }, \frac{c}{\sqrt{a^{2} +b^{2} +c^{2} } }$--------------(2)

substituting the values of a,b,c in equation 2.

The direction cosines we get from direction ratios are± $\frac{4}{\sqrt{4^{2} +(-3)^{2} +1^{2} } }$,±$\frac{3}{\sqrt{4^{2} +(-3)^{2} +1^{2} } }$,±$\frac{1}{\sqrt{4^{2} +(-3)^{2} +1^{2} } }$.

Thus direction cosines can be simplified to± $\frac{4}{\sqrt{29} }$,±$\frac{3}{\sqrt{29} }$,±$\frac{1}{\sqrt{29} }$.