Question

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

• A. $1$
• B. $2$
• C. $4$
• D. $3$

Answer 1

The correct option is A $1$

$x+y+2z-3=0=2x+3y+4z-4$ at $z-axis,x=y=0$

$x+y=3-2z$ and $2x+3y=4(1-z)$

solving $x$ and $y$ in function $z$,

$2x+3(3-2z-x)=4(1-z)$

$⟹2x+9-6z-3x=4-4z$

$⟹x=9-6z-4+4z$

$⟹x=5-2zequation\left(2\right)$

$⟹(5-2z)+y=3-2z$

$⟹y=3-2z-5+2z$

$y=-2equation\left(2\right)$

So, $\sqrt{x^2+y^2}=\sqrt{(-2{)}^2+(5-2z{)}^2}$

at, $z=\frac{5}{2}$, this would be minimum.