Browse Questions

# If from a point $P(a,b,c)$ perpendiculars $PA,\:PB,\:$ are drawn on $yz\:\:and\:\:zx$ planes. then the equation of the plane $OAB$ is ?

$(a)\:bcx+cay+abz=0\:\qquad\:(b)\:bcx+cay-abz=0\:\qquad\:(c)\:bcx-cay+abz=0\:\qquad\:(d)\:bcx-cay-abz=0$

Given that $A$ is foot of $\perp$ of $P(a,b,c)$ on $yz$ plane and
$B$ is foot of $\perp$ drawn from $P$ on z$x$ plane.
$\therefore \:A\:\:and\:\:B$ are given by $A(0,b,c)$ and $B(a,0,c)$
Normal to the plane $OAB$ is given by $(0,b,c)\times(a,0,c)= (bc,ca,-ab)$
$\therefore$ The eqn. of the plane with normal $(bc,ca,-ab)$ and passing through  $O(0,0,0)$ is
$bcx+cay-abz=0$