Question

The shortest distance between the $z$-axis and the line $x+y+2z-3=0=2x+3y+4z-4$, is :

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

The equation of any plane passing through given line is
$\left(x+y+2z-3\right)+\lambda \left(2x+3y+4z-4\right)=0$
$\Rightarrow \left(1+2\lambda \right)x+\left(x+3\lambda \right)y+\left(2+4\lambda \right)z-\left(3+4\lambda \right)=0$
If this plane is parallel to z-axis then normal to the plane will be perpendicular to z-axis.
$\therefore \left(1+2\lambda \right)\left(0\right)+\left(1+3\lambda \right)\left(0\right)\left(2+4\lambda \right)\left(1\right)=0$
$\lambda =-\frac{1}{2}$
Thus, Required plane is
$\left(x+y+2z-3\right)-\frac{1}{2}\left(2x+3y+4z-4\right)=0$
$\Rightarrow y+2=0$
$\therefore S.D=\frac{2}{\sqrt{{\left(1\right)}^2}}=2$