Discussion Forum

The equation of a plane passing through the line of intersection of the plane x+2y+3z=2 and x-y+z=3 and at a distance $2/\sqrt{3}$ from the point (3,1m-1) is 

Answer:

Explanation:

 Concept involved (i) equation of the plane through the intersection of two planes.

 i.e,   $(a_{1}x+b_{1}y+c_{1}z+d_{1})+\lambda (a_{2}x+b_{2}y+c_{2}z+d_{2})=0$

 (ii)  Distance of a point $(x_{1},y_{1},z_{1})$ from ax+by+cz+d=0

 $\Rightarrow$    $\frac{|ax_{1}+by_{1}+cz_{1}+d|}{\sqrt{a^{2}+b^{2}+c^{2}}}$

 Sol. Equation of plane passing through intersection of two planes

 $x+2y+3z=2$ and x-y+z=3 is 

$\Rightarrow$    (x+2y+3z-2)+$\lambda (x-y+z-3)=0$

$\Rightarrow$    $(1+\lambda)x+(2- \lambda)y+(3+\lambda)z-(2+3 \lambda)=0$

 whose distance from (3,1,-1) is    $\frac{2}{\sqrt{3}}$

   $\Rightarrow$    $\frac{ |3(1+\lambda )+1.(2-\lambda)-1(3+\lambda)-(2+3 \lambda)|}{\sqrt{(1+\lambda)^{2}+(2-\lambda)^{2}+(3+\lambda})^{2}}=\frac{2}{\sqrt{3}}$

  $\Rightarrow$  $\frac{|-2 \lambda|}{\sqrt{3 \lambda^{2}+4 \lambda+14}}=\frac{2}{\sqrt{3}}$

 $\Rightarrow$  $3 \lambda^{2}=3 \lambda^{2}+4 \lambda+14 \Rightarrow \lambda =-\frac{7}{2}$

$\therefore$   $\left(1-\frac{7}{2}\right)x+\left(2+\frac{7}{2}\right)y+\left(3-\frac{7}{2}\right)z-\left(2-\frac{21}{2}\right)=0$

 $\Rightarrow$    $-\frac{5x}{2}+\frac{11}{2}y-\frac{1}{2} z+\frac{17}{2}=0$

 or    5x+11y+z-17=0