Equation of a line is $\frac{x-3}{2}=\frac{4-y}{3}=\frac{2-z}{4}$ then its direction cosines are :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $\frac{2}{\sqrt{29}}, \frac{-3}{\sqrt{29}},\frac{-4}{\sqrt{29}}$
first converting eq. to std. form
$\frac{x-3}{2}=\frac{4-y}{3}=\frac{2-z}{4}$
DR's = 2, -3, -4
DC's = $\frac{2}{\sqrt{2^2+(-3)^2+(-4)^2}},\frac{-3}{\sqrt{2^2+(-3)^2+(-4)^2}},\frac{-4}{\sqrt{2^2+(-3)^2+(-4)^2}}$
DC's = $\frac{2}{\sqrt{29}}, \frac{-3}{\sqrt{29}},\frac{-4}{\sqrt{29}}$