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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  3-D Geometry
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The equation of the line through the point $(1,1,3)$ and parallel to the planes $\overrightarrow r.(\hat i-\hat j+2\hat k)=5$ and $\overrightarrow r.(3\hat i+\hat j+\hat k)=6$ is ?

$\begin{array}{1 1} (a)\:\:\overrightarrow r=(\hat i+\hat j+3\hat k)+\lambda(-3\hat i+5\hat j+4\hat k)\:\:& \:\:(b)\:\:\overrightarrow r=(-3\hat i+5\hat j+4\hat k)+\lambda(\hat i+\hat j+3\hat k)\\\:(c)\:\:\overrightarrow r=(-\hat i-\hat j-3\hat k)+\lambda(-3\hat i+5\hat j+4\hat k)\:\:&\:\:(d)\:\:\overrightarrow r=(3\hat i-5\hat j-4\hat k)+\lambda(\hat i+\hat j+3\hat k) \end{array} $

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  • $(\overrightarrow a\times\overrightarrow b)$ is a line (vector) $\perp$ to both $\overrightarrow a$ and $\overrightarrow b$.
Let $(l,m,n)$ be the $d.r.$ of the line.
Given that the line is parallel to the planes $\overrightarrow r.(\hat i-\hat j+2\hat k)=5$ and $\overrightarrow r.(3\hat i+\hat j+\hat k)=6$
$\Rightarrow$ The line is $\perp$ to normasl to both the planes.
$\therefore$ $(l,m,n)$ is along $\overrightarrow n_1\times\overrightarrow n_2$
$(l,m,n)=\left |\begin{array} {ccc} \hat i & \hat j & \hat k\\1 & -1 & 2\\ 3 & 1 & 1\end {array}\right |=(-3,5,4)$
$\therefore$ The eqn. of the line through $(1,1,3)$ and parallel to the two given planes is
$\overrightarrow r=(\hat i+\hat j+3\hat k)+\lambda(-3\hat i+5\hat j+4\hat k)$
answered Dec 31, 2013 by rvidyagovindarajan_1
 

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