Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Given equations: $\begin{array}{rl}& 2x+3y+z=1\\ & 2x+3y=1-z...(i)\\ & x+3y+2z=2\\ & x+3y=2-2z...(ii)\end{array}$ Subtracting (ii) from (i), we get $\begin{array}{rl}& 2x+3y-x-3y=1-z-2+2z\\ & x=-1+z\\ & z=\frac{x+1}{1}...(iii)\end{array}$ Putting value of $x$ in equation (ii), we get $\begin{array}{rl}& -1+z+3y=2-2z\\ & 3z=3-3y\\ & z=\frac{y-1}{-1}...(iv)\end{array}$ From (iii), (iv) $\frac{x+1}{1}=\frac{y-1}{-1}=\frac{z}{1}$ Thus, angle between above line and $\mathrm{X}$-axis having Direction Ratio's $(1,0,0)$ is given as $\begin{array}{rl}\cos \alpha & =\left|\frac{1\cdot (1)+0+0}{\sqrt{1+1+1}\cdot \sqrt{1}}\right|=\frac{1}{\sqrt{3}}\\ \therefore \sec \alpha & =\sqrt{3}\end{array}$