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Home  >>  TN XII Math  >>  Vector Algebra
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Find the vector and cartesian equation to the plane through the point $(-1 , 3 , 2 ) $ and perpendicular to the planes $x+2y+2z=5$ and $3x+y+2z=8.$

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  • Equation of the plane through a point $A(x_1, y_1, z_1)(pv \: is \: \overrightarrow a)$ and parallel to vectors $ \overrightarrow u$ and $\overrightarrow v$ $ \overrightarrow r=\overrightarrow a+s\overrightarrow u+t\overrightarrow v$ where $t,s$ are scalars (vector form) $ \begin{vmatrix}x-x_1 & y-y_1 & z-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$ (cartesian form) where $\overrightarrow u=l_1\overrightarrow i+m_1\overrightarrow j+n_1\overrightarrow k, \: \overrightarrow v=l_2\overrightarrow i+m_2\overrightarrow j+n_2\overrightarrow k$ Non-para meter vector equation : $ [ \overrightarrow r\: \overrightarrow u \: \overrightarrow v]=[\overrightarrow a\: \overrightarrow u\: \overrightarrow v]$
  • The direction ratios of the normal to the plane $ ax+by+cz=q$ (i.e., . $ \overrightarrow r.(a\overrightarrow i+b\overrightarrow j+c\overrightarrow k)=q)$ are $(a, b, c)$
Step 1
The required plane passes through the point $A(-1, 3, 2)(\overrightarrow a=-\overrightarrow i+3\overrightarrow j+2\overrightarrow k)$. It is perpendicular to the planes.
$ x+2y+2z=5\: and \: 3x+y+2z=8$
$ \therefore $ it is parallel to the normals to these two planes which are, respectively,
$ \overrightarrow u=\overrightarrow i+2\overrightarrow j+2\overrightarrow k\: and \: \overrightarrow v=3\overrightarrow i+\overrightarrow j+2\overrightarrow k$
Step 2
The vector equation of the plane is
$ \overrightarrow r=\overrightarrow a+s\overrightarrow u+t\overrightarrow v$
$\overrightarrow r=(-\overrightarrow i+3\overrightarrow j+2\overrightarrow k)+s(\overrightarrow i+2\overrightarrow j+2\overrightarrow k)+t(3\overrightarrow i+\overrightarrow j+2\overrightarrow k)$
Step 3
The cartesian equation is
$ \begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$
where $ (x_1, y_1, z_1)=(-1, 3, 2), (l_1, m_1, n_1 )=(1,2,2), (l_2, m_2, n_2)=(3, 1, 2)$
$ \begin{vmatrix} x+1 & y-3 & z-2 \\ 1 & 2 & 2 \\ 3 & 1 & 2 \end{vmatrix}=0$
$(4-2)(x+1)-(2-6)(y-3)+(1-6)(6-2)=0$
$ 2x+2+4y-12-5z+10=0$
$ 2x+4y-5z=0$

 

answered Jun 14, 2013 by thanvigandhi_1
edited Jun 25, 2013 by thanvigandhi_1
 

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