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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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True or False: The angle between the line $\overrightarrow{r}=(5\hat i-\hat j-\hat k)+\lambda(2\hat i-\hat j\;+\hat k)$ and the plane $\overrightarrow{r}.(3\hat i-4\hat j-\hat k)-5=0$ is $\sin^{-1}\large \frac{5}{2\sqrt 91}.$

$\begin{array}{1 1} True \\ False \end{array} $

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  • Angle between the line and the plane is $ \sin \theta= \large\frac{\overrightarrow b. \overrightarrow n}{|\overrightarrow b|.|\overrightarrow n|}$
Equation of the line $\overrightarrow{r}.(5\hat i-\hat j-\hat k)=(2\hat i-\hat j\;+\hat k)$
Equation of the plane $\overrightarrow{r}.(3\hat i-4\hat j-\hat k)=5$
Angle between the line and the plane is
$ \sin \theta=\large \frac{\overrightarrow b. \overrightarrow n}{|\overrightarrow b|.|\overrightarrow n|}$
Here $\overrightarrow{b}=2 \hat i-\hat j+\hat k$ and $\overrightarrow{n}=3 \hat i-4 \hat j-\hat k$
Now substituting this value in $\sin \theta$ we get,
$\sin \theta=\large\frac{(2 \hat i-\hat j+\hat k).(3 \hat i-4\hat j-\hat k)}{\sqrt {2^2+(-1)^2+1^2} \sqrt {3^2+(-4)^2+(-1)^2}}$
Applying the dot product we get,
$\sin \theta=\large\frac{6+4-1}{\sqrt 6\sqrt {26}}$
$\qquad=\large\frac{9}{\sqrt {156}}$
$\qquad=\large\frac{9}{2 \sqrt {39}}$
Hence the given statement is $False$
answered Jun 13, 2013 by meena.p
 

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