Discussion Forum

If a plane  passing through the point (2,2,1) and is perpendicular to the planes 3x+2y+4z+1=0 and  2x+y+3z+2=0 . Then the equation of the plane is 

Answer:

Explanation:

Equation of plane passing through (2,2,1) is

 a(x-2) +b(y-2)+c(z-1)=0   .....(i)

 Since , above plane is perpendicular to 

3x+2y+4z+1=0

 and 2x+y+3z+2=0

$\therefore$    3a+2b+4c=0       ......(ii)

 and 2a+b+3c=0     .....(iii)

 [ $\because$ for perpendicular , a₁a₂+b₁b₂+c₁c₂=0]

On multiplying eq.(iii) by 2 , we get

 4a+2b+6c=0     ......(iv)

 On subtracting eq.(iv) from eq.(ii) , we get

$\Rightarrow  c=\frac{-a}{2}$

 On putting $c=\frac{-a}{2}$ in eq. (iii) , we get b = $\frac{-a}{2}$

 on putting b = $\frac{-a}{2}$ and $c=\frac{-a}{2}$ in eq.(i)

we get   $a(x-2)-\frac{a}{2}(y-2)-\frac{a}{2}(z-1)=0$

$\Rightarrow\frac{a}{2}[2(x-2)-(y-2)-(z-1)]=0$

$\Rightarrow2x-4+y+2-z+1=0$

$\Rightarrow2x-y-z-1=0$