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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
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If $\overrightarrow a,\overrightarrow b,\overrightarrow c$ are position vectors of three non collinear points $A,B,C$ respectively, then the shortest distance of $A$ from $BC$ is ?

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

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  • Shortest distance of a point from a line is its $\perp$ distance.
  • Projection of $\overrightarrow a$ on $\overrightarrow b$ is $ \large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow b|}$
Vector equation of the line $BC$ is $\overrightarrow r=\overrightarrow b+\lambda(\overrightarrow c-\overrightarrow b)$
Let $P$ be the foot of $\perp$ of the point $A$ on $BC$.
Then the required distance is $AP$
Complete the rt. angled $\Delta\:ABP$
Then using pythogorous theorem, $AP=\sqrt {AB^2-BP^2}$
$AB^2=|\overrightarrow b-\overrightarrow a|^2$
$BP$ is projection of the vector $\overrightarrow {BA}$ on the line $\overrightarrow {BC}$
$i.e.,$ $BP=\large\frac{(\overrightarrow a-\overrightarrow b).(\overrightarrow c-\overrightarrow b)}{|\overrightarrow c-\overrightarrow b|}$
$\therefore$ The required distance $AP=\sqrt {|\overrightarrow b-\overrightarrow a|^2-\bigg(\large\frac{(\overrightarrow a-\overrightarrow b).(\overrightarrow c-\overrightarrow b)}{|\overrightarrow c-\overrightarrow b|}\bigg)^2}$
answered Jan 5, 2014 by rvidyagovindarajan_1

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