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The line joining the points $6\overrightarrow a-4\overrightarrow b-5\overrightarrow c$ and $-4\overrightarrow c$ and the line joining the points $-\overrightarrow a-2\overrightarrow b-3\overrightarrow c$ and $\overrightarrow a+2\overrightarrow b-5\overrightarrow c$ Intersects at the point ?

$\begin{array}{1 1} (a)\:2\overrightarrow c\:\:\:\qquad\:(b)\:-4\overrightarrow c\:\:\:\qquad\:(c)\:8\overrightarrow c\:\:\:\qquad\:(d)\:4\overrightarrow c. \end{array} $

1 Answer

  • Eqn. of a line joining the points $\overrightarrow a\:\:and\:\:\overrightarrow b$ is $\overrightarrow r=\overrightarrow a+\lambda (\overrightarrow a-\overrightarrow b)$
Equation of the line joining $6\overrightarrow a-4\overrightarrow b-5\overrightarrow c\:\;and\:\:-4\overrightarrow c$ is given by
$\overrightarrow r=-4\overrightarrow c+\lambda (6\overrightarrow a-4\overrightarrow b-\overrightarrow c)$.....(i)
Similarly eqn. of the line joining $-\overrightarrow a-2\overrightarrow b-3\overrightarrow c\:\:and\:\:\overrightarrow a+2\overrightarrow b-5\overrightarrow c$ is
$\overrightarrow r=(\overrightarrow a+2\overrightarrow b-5\overrightarrow c)+\mu(2\overrightarrow a+4\overrightarrow b-2\overrightarrow c)$ or
$\overrightarrow r=(\overrightarrow a+2\overrightarrow b-5\overrightarrow c)+\mu(\overrightarrow a+2\overrightarrow b-\overrightarrow c)$ .......(ii)
(i) and (ii) intersect if
$-4\overrightarrow c+\lambda (6\overrightarrow a-4\overrightarrow b-\overrightarrow c)=(\overrightarrow a+2\overrightarrow b-5\overrightarrow c)+\mu(\overrightarrow a+2\overrightarrow b-\overrightarrow c)$
$\Rightarrow\:(6\lambda \overrightarrow a-4\lambda \overrightarrow b-(4+\lambda)\overrightarrow c=(\mu+1)\overrightarrow a+(2\mu+2)\overrightarrow b+(-\mu-5)\overrightarrow c$
Equating the like terms on both the sides we get
Solving which we get $\lambda=0\:\:and\:\:\mu=-1$
$\therefore$ The point of intersection is got by substituting $\lambda$ in (i).
$\overrightarrow =-4\overrightarrow c$
answered Jan 5, 2014 by rvidyagovindarajan_1

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