Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equations of the line are given as $x=4z+3$ and $y=2-3z$.

Rewriting these equations in symmetric form:

$\frac{x-3}{4}=z$

$\frac{y-2}{-3}=z$

Combining them, we get:

$\frac{x-3}{4}=\frac{y-2}{-3}=\frac{z-0}{1}$

The direction ratios of the line are $4,-3,1$.

The direction cosines $\cos \alpha ,\cos \beta ,\cos \gamma$ are given by dividing the direction ratios by their magnitude $\sqrt{4^2+(-3{)}^2+1^2}=\sqrt{16+9+1}=\sqrt{26}$.

$\cos \alpha =\frac{4}{\sqrt{26}}$

$\cos \beta =\frac{-3}{\sqrt{26}}$

$\cos \gamma =\frac{1}{\sqrt{26}}$

Summing these values:

$\cos \alpha +\cos \beta +\cos \gamma =\frac{4-3+1}{\sqrt{26}}=\frac{2}{\sqrt{26}}$

Answer: $\frac{2}{\sqrt{26}}$