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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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True or False: The angle between the planes $\overrightarrow{r}.(2\hat i-3\hat j+\hat k)=1$ and $\overrightarrow{r}.(\hat i- j)=4$ is $\cos^{-1}\Large \frac{-5}{\sqrt {58}}.$

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

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Toolbox:
  • Angle between two planes
  • $ \theta= \cos ^{-1}\bigg[ \large\frac{\overrightarrow n_1. \overrightarrow n_2}{|\overrightarrow n_1|.|\overrightarrow n_2|} \bigg]$
Equation of the plane $P_1= \overrightarrow{r}.(2\hat i-3\hat j+\hat k)=1$
Equation of the plane $P_2= \overrightarrow{r}.(\hat i-\hat j)=4$
Angle between the two plane is given by
$ \theta= \cos ^{-1} \bigg[ \large\frac{\overrightarrow n_1. \overrightarrow n_2}{|\overrightarrow n_1|.|\overrightarrow n_2|} \bigg]$
Here $\overrightarrow{n_1}=2 \hat i-3\hat j+\hat k$
$|\overrightarrow{n_1}|=\sqrt {2^2+(-3)^2+1^2}$
$\qquad=\sqrt {4+9+1}$
$\qquad=\sqrt {14}$
$\overrightarrow{n_2}= \hat i-\hat j$
$|\overrightarrow{n_2}|=\sqrt {1^2+(-1)^2}$
$\qquad=\sqrt {2}$
Now substituting the values we get,
$\cos \theta=\large\frac{(2 \hat i-3\hat j+\hat k).( \hat i-\hat j)}{\sqrt {14} \sqrt {2}}$
Applying the dot product we get,
$\cos \theta=\large\frac{2+3}{\sqrt {28}}$
$\cos \theta=\large\frac{5}{2 \sqrt {7}}$
Hence the statement is $False$
answered Jun 13, 2013 by meena.p
 

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