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Home  >>  TN XII Math  >>  Vector Algebra
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Find the equations of the plane which contains the two lines $\large\frac{x+1}{2}=\frac{y-2}{-3}=\frac{z-3}{4}$ and $\large\frac{x-4}{3}=\frac{y-1}{2}=z-8$

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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]$
Step 1
The plane contains the lines $ \large\frac{x+1}{2}=\large\frac{y-2}{-3}=\large\frac{z-3}{4}\: and \: \large\frac{x-4}{3}=\large\frac{y-1}{2}=z-8$
$ \therefore $ it contains the point $A(-1, 2, 3) ( \overrightarrow a=-\overrightarrow i+2\overrightarrow j+3\overrightarrow k)$ which lies on the first line and it is parallel to the vectors
$ \overrightarrow u=2\overrightarrow i-3\overrightarrow j+4\overrightarrow k $ ( parallel to the first line) and
$ \overrightarrow v=3\overrightarrow i+2\overrightarrow j+\overrightarrow k$ ( parallel to the second line )
If $ \overrightarrow r $ is the pv of any point on the plane, the vectors $ \overrightarrow r- \overrightarrow a, \: \overrightarrow u\: and \: \overrightarrow v$ are coplanar.
Step 2
Therefore the non-parametric vector equation of the plane is $[ \overrightarrow r-\overrightarrow a\: \: \overrightarrow u\: \overrightarrow v]=0 \: or \: [\overrightarrow r-(-\overrightarrow i+2\overrightarrow j+3\overrightarrow k)\:\: \: \: \: 2\overrightarrow i-3\overrightarrow j+4\overrightarrow k\:\: \: \: 3\overrightarrow i+2\overrightarrow j+\overrightarrow k]=0$
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 $(x_1, y_1, z_1)=(-1, 2, 3), (l_1, m_1, n_1 )=(2, -3, 4), (l_2, m_2, n_2)=(3, 2, 1)$
$ \begin{vmatrix} x+1 & y-2 & z-3 \\ 2 & -3 & 4 \\ 3 & 2 & 1 \end{vmatrix}=0$
$(x+1)(-3-8)-(y-2)(2-12)+(z-3)(4+9)=0$
$ -11x-11+10y-20+13z-39=0$
$ -11x+10y+13z=70$ is the equation of the plane.

 

answered Jun 16, 2013 by thanvigandhi_1
edited Jun 25, 2013 by thanvigandhi_1
 

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