Question

Given $A(1,-1,0)$; $B(3,1,2)$; $C(2,-2,4)$ and $D(-1,1,-1)$ which of the following points neither lie on $AB$ nor on $CD$?

• A. $(2,2,4)$
• B. $(2,-2,4)$
• C. $(2,0,1)$
• D. $(0,-2,-1)$

Answer 1

The correct option is A $(2,2,4)$

$A(1,-1,0),B(3,1,2),C(2,-2,4),D(-1,1,-1)\overrightarrow{AB}=<3-1,1-(-1),2-0>$

$=<2,2,2>$

$\therefore EquationoflineAB:\frac{x-1}{2}=\frac{y-(-1)}{2}=\frac{z-0}{2}⟹\frac{x-1}{2}=\frac{y+1}{2}=\frac{z-0}{2}$

$\overrightarrow{CD}=<-1-2,1-(-2),-1-4>$

$=<-3,3,-5>$

$\therefore EquationoflineCD:\frac{x-2}{-3}=\frac{y-(-2)}{3}=\frac{z-4}{-5}⟹\frac{x-2}{-3}=\frac{y+2}{3}=\frac{z-4}{-5}$

By putting the values of points given in the choices in the equation of lines $AB$ and $CD$, $(2,2,4)$ is the point which neither lie on $AB$ nor on $CD$.