equation of plane passes through \( (4,2,1) \) and \( (2,1,-1) \) making angles of \( 45^0 \) with the plane \( x-4y+z-9=0 \)
Express the line \( r \) through \( (4,2,1) \) and \( (2,1,-1) \) as intersection of two planes \( \pi\equiv ax+by+cz+d=0 \) and \( \pi'\equiv a'x+b'y+c'z+d'=0 \). The equation of all planes containing \( r \) is \( \pi_{\lambda}\equiv ax+by+cz+d+\lambda (a'x+b'y+c'z+d')=0 \) (\( \pi' \) excluded). Now, use \( \cos 45^0=\displaystyle\frac{\sqrt[ ]{2}}{2}=\displaystyle\frac{v_{\pi_{\lambda}}\cdot (1,-4,1)}{ \left\|{v_{\pi_{\lambda}}}\right\| \left\|{(1,-4,1)}\right\|} \) and you'll obtain \( \lambda \).
P.S. Please change the title of this thread.
