Question

The plane $4x+7y+4z+81=0$ is rotated through a right angle about its line of intersection with the plane $5x+3y+10z=25.$ If the equation of the plane in its new position is $x-4y+6z=K,$ then the value of $K$ is

• A. $106$
• B. $-89$
• C. $73$
• D. $37$

Answer 1

Answer: (A)

The equation of the plane through the line of intersection of the planes $\left(4x+7y+4z+81\right)+\lambda \left(5x+3y+10z-25\right)=0$
$\left(4+5\lambda \right)x+\left(7+3\lambda \right)y+\left(4+10\lambda \right)z+\left(81-25\lambda \right)=0$
Which is parallel to $x-4y+6z=K$
$\Rightarrow \frac{4+5\lambda }{1}=\frac{7+3\lambda }{-4}=\frac{4+10\lambda }{6}=\frac{\left(81-25\lambda \right)}{-K}$
So, $-16-20\lambda =7+3\lambda$
$23\lambda =-23$
$\lambda =-1$
and $\frac{81-25\lambda }{-K}=\frac{4+10\lambda }{6}$
$-K=\frac{\left(81-25\lambda \right)6}{\left(4+10\lambda \right)}=-106$
Hence, $K=106$