Question

If the image of the point $(1,1,1)$ by a plane is $(3,-1,5)$ then the equation of the plane is $ax+by+cz+d=0$ then

• A. $d<b<a<c$
• B. $d<a<b<c$
• C. $b<d<c<a$
• D. $c<d<a<b$

Answer 1

Answer: (A)

$d<b<a<c$

$\frac{3-1}{a}=\frac{-1-1}{b}=\frac{5-1}{c}=-2\frac{(a+b+b+c+d)}{a^2+b^2+c^2}$
$\frac{2}{a}=\frac{-2}{b}=\frac{4}{c}=\frac{-1(a+b+c+d)}{a^2+b^2+c^2}$
$\frac{-1}{a}=\frac{+1}{b}=\frac{-2}{c}=\frac{a+b+b+d}{a^2+b^2+c^2}=\frac{1}{\lambda }$
$a=-\lambda$, $b=\lambda$, $c=-2\lambda$
$a^2+b^2+c^2=a\lambda +b\lambda +c\lambda +d\lambda$
${\lambda }^2+{\lambda }^2+4\lambda =-{\lambda }^2+{\lambda }^2-2{\lambda }^2+d\lambda$
$6{\lambda }^2=-2{\lambda }^2+d\lambda$
$8{\lambda }^2=d\lambda \Rightarrow d=8\lambda$
If $\lambda >0$
$c<a<b<d$
If $\lambda <0$
$d<b<a<c$