Question

The plane $x-y-z=2$ is rotated through an angle ${90}^o$ about its line of intersection with the plane $x+2y+z=2$. Then equation of this plane in new position is

• A. $5x+4y+z-10=0$
• B. $4x+5y-3z=0$
• C. $2x+y+2z=9$
• D. $3x+4y-5z=9$

Answer 1

The correct option is A $5x+4y+z-10=0$

According tp the question,

$\begin{array}{c}if,p_1:x-y-z+2=0\\ p_2:x+2y+z-2=0\\ Newplane,p_1+\lambda \left(p_2\right)=0\\ \Rightarrow (x-y-z+2)+\lambda (x+2y+z-2)=0\\ \Rightarrow x-y-z-2+x\lambda +2y\lambda +z\lambda -2\lambda =0\\ \therefore p_3:x(1+\lambda )+y(-1+2\lambda )+z(-1+\lambda )-2(1+\lambda )=0---\left(1\right)\\ Now,\\ plane{p}_1\&p_{3}areperpendiculartoeachother.\\ a_1{a}_2+b_1{b}_2+c_1{c}_2=0\\ \Rightarrow \left(1\right)(1+\lambda )+(-1)(-1+2\lambda )+(-1)(-1+\lambda )=0\\ \Rightarrow 1+\lambda +1-2\lambda +1-\lambda )=0\\ \Rightarrow 3-2\lambda =0\\ \therefore \lambda =\frac{3}{2}\\ lets,\\ putthevalueof\lambda ineq{u}^n......\left(1\right)\\ \Rightarrow x(1+\lambda )+y(-1+2\lambda )+z(-1+\lambda )-2(1+\lambda )=0\\ \Rightarrow x(1+\frac{3}{2})+y(-1+2\times \frac{3}{2})+z(-1+\frac{3}{2})-2(1+\frac{3}{2})=0\\ \Rightarrow \frac{5}{2}x+2y+\frac{1}{2}z-5=0\\ \Rightarrow \frac{5x+4y+z-10}{2}=0\\ \therefore 5x+4y+z-10=0\\ sothatthenewpositionofplaneis:5x+4y+z-10=0\\ and,thecorrectoptionisA.\end{array}$