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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
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If the distance of the point $P(1,-2,1)$ from the plane $x+2y-2z=a$, where $a>0,$ is $5$, then the foot of the $\perp$ from $P$ to this plane is ?

$\begin{array}{1 1}\:(a)\:\:\big(\large\frac{8}{3},\frac{4}{3},-\frac{7}{3}\big)\:\:\:\qquad\:\:(b)\:\:\big(\large\frac{4}{3},-\frac{4}{3},-\frac{1}{3}\big)\:\:\:\qquad\:\:(c)\:\:\big(\large\frac{1}{3},\frac{2}{3},-\frac{10}{3}\big)\:\:\qquad\:\:(d)\:\:\big(\large\frac{2}{3},-\frac{1}{3},-\frac{5}{3}\big) \end{array} $

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  • Distance of the point $(x_1,y_1,z_1)$ from the plane $ax+by+cz+d=0$ is $\bigg|\large\frac{ax_1+by_1+cz_1+d}{\sqrt {a^2+b^2+c^2}}\bigg|$
  • If two lines (vectors) are parallel, then their $d.r.^s$ are proportional.
Given that the distance of $P(1,-2,1)$ from the plane $x+2y-2z-a=0$ is $5$
$\therefore\:\bigg|\large\frac{1-4-2-a}{\sqrt {1+4+4}}\bigg|=5$
But since $a>0,\:\:a=10$
$\therefore\:$ The eqn. of the plane is $x+2y-2z-10=0$
$\Rightarrow\:d.r.$ of normal to the plane $\overrightarrow n$ is $(1,2,-2)$
Let The foot of $\perp$ from $P$ on the plane be $Q(x_1,y_1,z_1)$
$d.r.$ of $\overline {PQ}$ is $ (x_1-1,y_1+2,z_1-1)$
Since $\overline {PQ}$ is $\perp$ to the plane, $\overline {PQ}$ is $||$ to the normal $\overrightarrow n$ to the plane.
But since $Q$ lies on the plane, it satisfies the eqn. of the plane.
$\therefore$ The foot of $\perp$ is $Q\big(\large\frac{8}{3},\frac{4}{3},-\frac{7}{3}\big)$
answered Dec 27, 2013 by rvidyagovindarajan_1

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