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Questions  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  3-D Geometry

Distance of a point $P(\overrightarrow c)$ from the the line $\overrightarrow r=\overrightarrow a+\lambda \overrightarrow b$ is ?

$\large (a)\:\bigg|\frac{(\overrightarrow c-\overrightarrow a)\times\overrightarrow b}{|\overrightarrow b|}\bigg|\:\qquad\:(b)\:\bigg|\frac{(\overrightarrow c-\overrightarrow a).\overrightarrow b}{|\overrightarrow b|}\bigg|\:\qquad\:(c)\:\bigg|\frac{(\overrightarrow c-\overrightarrow a)\times\overrightarrow b}{|\overrightarrow b|^2}\bigg|\:\qquad\:(d)\:None\:of\:these.$

1 Answer

  • $|\overrightarrow a\times\overrightarrow b|^2=|\overrightarrow a|^2 |\overrightarrow b|^2-(\overrightarrow a .\overrightarrow b)^2$
Let the foot of $\perp$ from $P$ on the given line $\overrightarrow r=\overrightarrow a+\lambda \overrightarrow b$ be Q.
$\Rightarrow\: A(\overrightarrow a)$ and $Q$ are two points on the line.
$\therefore \overrightarrow {AP}=\overrightarrow c-\overrightarrow a$
$AQ$ is projection of $AP$ on the given line $(\overrightarrow b)$
$i.e., \:AQ=\large\frac{\overrightarrow {AP}.\overrightarrow {b}}{|\overrightarrow b|}$
From the rt. angled $\Delta\: APQ$,using pythogorous theorem,
The required distance $PQ=\sqrt {AP^2-AQ^2}$
$=\sqrt {|\overrightarrow c-\overrightarrow a|^2-\large\frac{|(\overrightarrow c-\overrightarrow a).\overrightarrow b|^2}{|\overrightarrow b|^2}}$
$=\sqrt {\large\frac{|\overrightarrow c-\overrightarrow a|^2.|\overrightarrow b|^2-|(\overrightarrow c-\overrightarrow a).\overrightarrow b|^2}{|\overrightarrow b|^2}}$
But $ |\overrightarrow c-\overrightarrow a|^2.|\overrightarrow b|^2-|(\overrightarrow c-\overrightarrow a).\overrightarrow b|^2=|(\overrightarrow c-\overrightarrow a)\times\overrightarrow b|^2$
$\therefore\:PQ=\large\frac{(\overrightarrow c-\overrightarrow a)\times\overrightarrow b}{|\overrightarrow b|}$
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