Question

Find the equation of the plane which is perpendicular to the plane 5x+3y+6z+8=0 and which contains the line of intersection of the planes x+2y+3z-4=0 and 2x+y-z+5=0.

Answer 1

The equation of the plane through the line of intersection of the planes x+2y+3z-4=0 and 2x+y-z+5=0 is

$(x+2y+3z-4=0)+\lambda (2x+y-z+5=0)=0\Rightarrow x(1+2\lambda )+y(2+\lambda )+z(-\lambda +3)-4+5\lambda =0...\left(i\right)Also,thisisperpendiculartotheplane5x+3y+6z+8=0.\therefore 5(1+2\lambda )+3(2+\lambda )+6(\lambda -3)=0[\because a_1{a}_2+b_1{b}_2+c_1{c}_2=0]\Rightarrow 5+10\lambda +6+3\lambda +18-6\lambda =0\therefore \lambda =\frac{-29}{7}FromEq.\left(i\right),x[1+2\left(\frac{-29}{7}\right)]+y(2-\frac{29}{7})+z(\frac{29}{7}+3)-4+5\left(\frac{-29}{7}\right)=0\Rightarrow x(7-58)+y(14-29)+z(29+21)-28-145=0\Rightarrow -51x-15y+50z-173=0So,therequiredequationofplaneis51x+15y-50z+173=0.$