Question

If the planes $x=cy+bz,y=az+cx$ and $z=bx+ay$ pass through a line then

• A. $a^2+b^2+c^2+2abc=0$
• B. $a^2+b^2+c^2+2abc=1$
• C. $a^2+b^2+c^2=2abc$
• D. $a+b+c=abc$

Answer 1

Answer: (B)

$a^2+b^2+c^2+2abc=1$

Given equation of the planes are

$x-cy-bz=\mathrm{0....}\left(i\right)$

$xx-y+az=\mathrm{0....}\left(ii\right)$

$bx+ay-z=\mathrm{0....}\left(iii\right)$

If a plane passes through a line then it is perpendicular to the normal of the plane,

Let the direction ratio of the line be (l,m,n)

Then,

$(l,m,n).(1,-c,-b)=0$, $(l,m,n).(c,-1,a)=0$ and $(l,m,n).(b,a,-1)=0$

$l-mc-nb=\mathrm{0.....}\left(A\right)$

$lc-m+na=\mathrm{0....}\left(B\right)$

$bl+am-n=\mathrm{0....}\left(C\right)$

Solving (A) and (B) by cross multiplication

$\frac{l}{-ac-b}=\frac{-m}{a+bc}=\frac{n}{-1+c^2}$

$\Rightarrow l=-ac-b$

$\Rightarrow m=-a-bc$

$\Rightarrow n=c^2-1$

Substituting the values in (C), we get

$b(-ac-b)+a(-a-bc)-c^2+1=0$

$\Rightarrow -2abc-a^2-b^2-c^2+1=0$

$\Rightarrow a^2+b^2+c^2+2abc=1$