Question

The equation of the plane which bisects the angle between the planes $3x-6y+2z+5=0$ and $4x-12y+3z-3=0$ which contains the origin is ?

• A. $33x-13y+32z+45=0$
• B. $x-3y+z-5=0$
• C. $33x+13y+32z+45=0$
• D. $Noneofthese$

Answer 1

Answer: (D)

$Noneofthese$

Given ,

The required equation of plane bisects the given two planes.

$\begin{array}{c}\therefore \frac{3x-6y+2z+5}{\sqrt{3^2+{\left(-6\right)}^2+{\left(2\right)}^2}}=\pm \frac{4x-12y+3z-3}{\sqrt{4^2+{\left(-12\right)}^2+3^2}}\\ \frac{3x-6y+2z+5}{\sqrt{49}}=\pm \frac{4x-12y+3z-3}{\sqrt{169}}\\ \frac{3x-6y+2z+5}{7}=\pm \frac{4x-12y+3z-3}{13}\\ 39x-78y+26z+65=\pm 28x-84y+21z-21\end{array}$

Now, solving for the positive value, we get

$39x-78y+26z+65$.........(i)

or, $11x+6y+5z+36=0$

And for negative value, we get

$\begin{array}{c}39x-78y+26z+65=-\left(28x-84y+21z-21\right)\\ or,67x-162y+47z+44=\mathrm{0.......}\left(ii\right)\end{array}$

$\because$None of the answer matches with the given equation.

Hence,

Option $D$ is correct in this case.