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Home  >>  TN XII Math  >>  Vector Algebra
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Find the vector and cartesian equations of the plane through the point $(1 , 3 , 2 ) $ and parallel to the lines $\large\frac{x+1}{2}=\frac{y+2}{-1}=\frac{z+3}{3}$ and parallel to the line $\large\frac{x-2}{1}=\frac{y+1}{2}=\frac{z+2}{2}$

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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 line $ \large\frac{x-x_1}{l}=\large\frac{y-y_1}{m}=\large\frac{z-z_1}{n}$ contains the point $(x_1, y_1, z_1) $ and is parallel to the vector $ l\overrightarrow i+m\overrightarrow j + n\overrightarrow k$
Step 1
The plane passes through the point $A(1, 3, 2) \large\frac{x+1}{2}=\large\frac{y+2}{-1}=\large\frac{z+3}{3} \: (\overrightarrow a=\overrightarrow i+3\overrightarrow j+2\overrightarrow k)$ and is parallel to the lines $ \large\frac{x-2}{1}=\large\frac{y+1}{2} = \large\frac{z=2}{2}$ which are respectively parallel to $\overrightarrow u = 2\overrightarrow i-\overrightarrow j+3\overrightarrow k$ and $\overrightarrow v=\overrightarrow i+2\overrightarrow j+2\overrightarrow k$
Step 2
The vector equation of the plane is of the form $ \overrightarrow r=\overrightarrow a+s\overrightarrow u+t\overrightarrow v$ i.e.,
$ \overrightarrow r=\overrightarrow i+3\overrightarrow j+2\overrightarrow k+s(2\overrightarrow i-\overrightarrow j+3\overrightarrow k)+t(\overrightarrow i+2\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 ) = (2, -1, 3), (l_2, m_2, n_2)=(1, 2, 2)$
$ \therefore \begin{vmatrix} x_1 & y-3 & z-2 \\ 2 & -1 & 3 \\ 1 & 2 & 2 \end{vmatrix}=0$
$ (x-1)(-2-6)-(y-3)(4-3)+(z-2)(4+1)=0$
$ -8x+8-y+3+5z-10=0$
$ 8x+y-5z=1$

 

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

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