Q. Perpendicular are drawn from points on the line $ \frac{x+2}{2}=\frac{y+1}{-1}= \frac{z}{3}$ to the plane $x+ y + z = 3$. The feet of perpendiculars lie on the line

Solution:

PLAN To find the foot of perpendiculars and find its locus.
$\, \, \, \, \, \, \, \, $ Formula used
$\, \, \, \, \, \, \, \, $ Foot of perpendicular from $(x_1,y_1,z_1)$ to
$\, \, \, \, \, \, \, \, $ $ax+by+cz+d = 0\, be \, (x_2,y_2,z_2)$ then
$ \hspace25mm \frac{x_2-x_1}{a} = \frac{y_2-y_1}{b} = \frac{z_2-z_1}{c}$
$ \hspace35mm = \frac{-(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2} $
Any point on $\frac{x+2}{2} = \frac{y+1}{-1} =\frac{z}{3} = \lambda$
$\Rightarrow$ $\hspace30mm x = 2 \lambda - 2, y = - \lambda - 1, z = 3 \lambda$
Let foot of perpendicular from $ (2 \lambda -2,-\lambda-1,3\lambda)$
to x+ y + z = 3 be $(x_2,y_2,z_2)$.
$\therefore$ $\, \, \, \frac{x_2-(2 \lambda-2)}{1} = \frac{y_2-(- \lambda-1)}{1} =\frac{z_2-(3 \lambda)}{1}$
$ \hspace30mm =\frac{(2 \lambda-2 - \lambda - 1 + 3\lambda-3 )}{1+1+1} $
$ \Rightarrow \, \, \, \, x_2 - 2 \lambda + 2= y_2 + \lambda + 1 = z_2 - 3\lambda = 2 -\frac{4\lambda}{3}$
$\therefore$ $ \hspace20mm x_2 =\frac{2 \lambda}{3}, y_2 = 1 - \frac{7 \lambda}{3}, + 2= z_2 = 2 + \frac{5 \lambda}{3} $
$\Rightarrow $ $\hspace20mm \lambda =\frac{x_2 - 0 }{2/3}, = 1 - \frac{y_2-1}{-7/3}, = + \frac{z_2 - 2}{5/3} $
Hence, foot of perpendicular lie on
$=\frac{x }{2/3} = \frac{y-1}{-7/3} = \frac{z - 2}{5/3} \, \, \, \Rightarrow \, \, \frac{x }{2}= \frac{y-1}{-7} = \frac{z - 2}{5} $

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