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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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Determine whether the given planes are parallel or perpendicular, and in case they are neither, then find the angles between them.$(e)\; 4x + 8y + z -8 = 0 $ and $ y + z -4 =0$

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

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  • The direction ratios of normal to the plane $l_1=a_1x+b_1y+c_1 x=0$ are $a_1,b_1$ and $c_1$ and $l_2=a_2x+b_2y+c_2z=0$ are $a_2,b_2,c_2$
  • The angle between the planes is given by $\cos\theta=\begin{vmatrix}\large\frac{\overrightarrow n_1.\overrightarrow n_2}{\mid\overrightarrow n_1\mid\mid\overrightarrow n_2\mid}\end{vmatrix}$
  • $\Rightarrow \begin{vmatrix}\large\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\end{vmatrix}$
Step 1:
The equation of the given planes are :
$4x+8y+z-8=0$ and $y+z-4=0$
The direction cosines of $L_1$ are (4,8,1)
The direction cosines of $L_2$ are (0,1,1)
Therefore $\cos\theta=\begin{vmatrix}\large\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\end{vmatrix}$
Step 2:
Substituting for $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$
$\cos\theta=\begin{vmatrix}\large\frac{4\times 0+8\times 1+1\times 1}{\sqrt{4^2+8^2+1^2}\sqrt{0^2+1^2+1^2}}\end{vmatrix}$
$\Rightarrow \begin{vmatrix}\large\frac{0+8+1}{\sqrt{16+64+1}\sqrt{0+1+1}}\end{vmatrix}$
$\Rightarrow \large\frac{9}{\sqrt{81}\sqrt{2}}$
$\Rightarrow \large\frac{9}{9\sqrt{2}}$
Therefore $\cos\theta=\large\frac{1}{\sqrt{2}}$
$\Rightarrow \theta=\large\frac{\pi}{4}$
answered Jun 3, 2013 by sreemathi.v
 

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