Question

The plane $x-2y+3z=0$ is rotated through a right angle about its line of intersection with the plane $2x+3y-4z=0$. The equation of the plane in its new position is-

• A. $22x+5y-4z=35$
• B. $x-8y+13z=35$
• C. $22x-4y+6z=110$
• D. $22x-5y+13z=105$

Answer 1

Answer: (A)

$22x+5y-4z=35$

Here we have to find the equation of the plane through the line of intersection of the planes $x-2y+3z=0$ and $2x+3y-3z+5=0$ and at right angles to the plane $x-2y+3z=0$.
Let the equation of the plane be
$x-2y+3z+\lambda (2x+3y-4z-5)=0\Rightarrow (1+2\lambda )x+(-2+3\lambda )y+(3-4\lambda )-5\lambda =0$ ...(1)
It this plane is perpendicular to the plane $x-2y+3z=0$, then
$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
i.e $1.(1+2\lambda )-2(3\lambda -2)+3(3-4\lambda )=0\Rightarrow \lambda =\frac{7}{8}$
Substituting the value of $\lambda$ in (1), we get
$22x+5y-4z=35$