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The angle between the line $\overrightarrow r=(2\hat i-\hat j+3\hat k)+\lambda(3\hat i-\hat j+2\hat k)\:\:and\:\:the\:\:plane\:\:\overrightarrow r.(\hat i+\hat j+\hat k)=3$ is ?

$\begin{array}{1 1} cos^{-1}\large\frac{4}{\sqrt {42}} \\ sin^{-1}\large\frac{4}{\sqrt {42}} \\ tan^{-1}\large\frac{4}{\sqrt {42}} \\ tan^{-1}4\end{array} $

1 Answer

  • Angle between the line $\overrightarrow r=\overrightarrow a+\lambda \overrightarrow b$ and the plane $\overrightarrow r.\overrightarrow n=k$ is $sin^{-1}\bigg(\large\frac{\overrightarrow b.\overrightarrow n}{|\overrightarrow b| |\overrightarrow n|}\bigg)$
Eqn. of the line is $\overrightarrow r=(2\hat i-\hat j+3\hat k)+\lambda(3\hat i-\hat j+2\hat k)$
$\Rightarrow\:\overrightarrow a=2\hat i-\hat j+3\hat k,\:\:and\:\:\overrightarrow b=3\hat i-\hat j+2\hat k$
Eqn. of the plane is $\overrightarrow r.(\hat i+\hat j+\hat k)=3$
$\Rightarrow\:\overrightarrow n=\hat i+\hat j+\hat k$
$\Rightarrow\:$ Angle between the line and the plane is $sin^{-1}\bigg(\large\frac{(3\hat i-\hat j+2\hat k).(\hat i+\hat j+\hat k)}{\sqrt {14}.\sqrt 3}\bigg)$
$=sin^{-1}\big(\large\frac{4}{\sqrt {42}}\big)$
answered Dec 18, 2013 by rvidyagovindarajan_1

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