Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The given equation of the line is $\frac{2x+5}{3}=\frac{2-3z}{-1}$ and $3y=-2$.

Rewriting the equations in the standard symmetric form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$, we get:

$\frac{2(x+\frac{5}{2})}{3}=\frac{-3(z-\frac{2}{3})}{-1}$ and $y=\frac{-2}{3}$

$\frac{x+\frac{5}{2}}{\frac{3}{2}}=\frac{y+\frac{2}{3}}{0}=\frac{z-\frac{2}{3}}{\frac{1}{3}}$

The direction ratios of the line are proportional to $\frac{3}{2},0,\frac{1}{3}$.

Multiplying by $6$, the direction ratios are $9,0,2$.

The magnitude is $\sqrt{9^2+0^2+2^2}=\sqrt{81+0+4}=\sqrt{85}$.

The direction cosines of the line are $\frac{9}{\sqrt{85}},0,\frac{2}{\sqrt{85}}$.

Comparing this with the given direction cosines $\frac{9}{\sqrt{85}},m,\frac{2}{\sqrt{85}}$, we get $m=0$.

Answer: $0$