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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Find the angle between the planes whose vector equations are $ \overrightarrow r. ( 2\hat i + 2\hat j - 3\hat k ) = 5 \: and\: \overrightarrow r. ( 3\hat i - 3\hat j + 5\hat k ) = 3. $

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Toolbox:
  • If $\overrightarrow {n_1}$ and $\overrightarrow {n_2}$ are normal to the planes $\overrightarrow r.\overrightarrow n_1=d_1$ and $\overrightarrow r.\overrightarrow n_2=d_2$,then the angle between them is given by
  • $\cos\theta=\begin{vmatrix}\large\frac{\overrightarrow n_1.\overrightarrow n_2}{\mid\overrightarrow n_1\mid.\mid\overrightarrow n_2\mid}\end{vmatrix}$
Step 1:
The equation of the given planes are :
$\overrightarrow r.(2\hat i+2\hat j-3\hat k)=5$ and $\overrightarrow r.(3\hat i-3\hat j+5\hat k)=3$
Here $\overrightarrow n_1=(2\hat i+2\hat j-3\hat k)$
$\quad\overrightarrow n_2=(3\hat i-2\hat j+5\hat k)$
$\mid\overrightarrow n_1\mid=\sqrt{(2)^2+(2)^2+(-3)^2}$
$\qquad\;=\sqrt{4+4+9}$
$\qquad\;=\sqrt{17}$
$\mid\overrightarrow n_2\mid=\sqrt{(3)^2+(-3)^2+(5)^2}$
$\qquad\;=\sqrt{9+9+25}$
$\qquad\;=\sqrt{43}$
Step 2:
We know that $\cos\theta=\begin{vmatrix}\large\frac{\overrightarrow n_1.\overrightarrow n_2}{\mid\overrightarrow n_1\mid.\mid\overrightarrow n_2\mid}\end{vmatrix}$
$\cos\theta=\begin{vmatrix}\large\frac{(2\hat i+2\hat j-3\hat k).(3\hat i-3\hat j+5\hat k)}{\sqrt{17}\sqrt{43}}\end{vmatrix}$
$\qquad=\begin{vmatrix}\large\frac{6-6-15}{\sqrt{17}\sqrt{43}}\end{vmatrix}$
Step 3:
On simplifying we get
$\cos\theta=\large\frac{15}{\sqrt{17}\sqrt{43}}$
$\Rightarrow \theta=\cos^{-1}\big(\large\frac{15}{\sqrt{731}}\big)$
Hence the angle between the two plane is $\theta=\cos^{-1}\big(\large\frac{15}{\sqrt{731}}\big)$
answered Jun 3, 2013 by sreemathi.v
 

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