Question

Direction cosines of the line $\frac{x+2}{2}=\frac{2y-5}{3},z=-1$ are ____

• A. $\frac{4}{5},\frac{3}{5},0$
• B. $\frac{3}{5},\frac{4}{5},\frac{1}{5}$
• C. $-\frac{3}{5},\frac{4}{5},0$
• D. $\frac{4}{5},-\frac{2}{5},\frac{1}{5}$

Answer 1

Answer: (A)

$\frac{4}{5},\frac{3}{5},0$

$\frac{x+2}{2}=\frac{2y-5}{3},z=-1$

$\Rightarrow$ $\frac{x+2}{4}=\frac{y-2.5}{3},z=-1$

$\therefore$ direction cosine in $x$ $($ $\alpha$ $)=4k$, direction cosine in $y$ $($ $\beta$ $)$ $=3k$, direction cosine in $z($ $\gamma$ $)=0$

But ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

Therefore, $16k^2+9k^2=1$

$\Rightarrow 25k^2=1$

$\Rightarrow k=\frac{1}{5}$

Therefore, direction cosines of the line are $(\frac{4}{5},\frac{3}{5},0)$.