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Home  >>  TN XII Math  >>  Vector Algebra
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Find the vector cartesian equation of the plane passing through the points with position vectors $\overrightarrow{3i}+\overrightarrow{4j}+\overrightarrow{2k}, \overrightarrow{2i}-\overrightarrow{2j}-\overrightarrow{2k},$ and $\overrightarrow{7i}+\overrightarrow{k}.$

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  • Vector equation of the plane through three noncollinear points $A(x_1, y_1, z_1)(pv\: \overrightarrow a), B(x_2, y_2, z_2) (pv\: \overrightarrow b)\: and \: C(x_3, y_3, z_3)(pv\: \overrightarrow c)$ $ \overrightarrow r=(1-s-t)\overrightarrow a+s\overrightarrow b+t\overrightarrow c $ (vector equation)
  • $ \begin{vmatrix} x-x_1 & y_2-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0 $ (Cartesian equation )
  • $ [ \overrightarrow r-\overrightarrow a\:\: \overrightarrow b-\overrightarrow a\: \: \overrightarrow c-\overrightarrow a]=0 $ ( nonparametric vector equation )
Step 1
The plane passes through the points A, B, C whose position vector are $ \overrightarrow a = 3\overrightarrow i+4\overrightarrow j+2\overrightarrow k, \: \overrightarrow b=2\overrightarrow i-2\overrightarrow j-\overrightarrow k\: \overrightarrow c=7\overrightarrow i+\overrightarrow k.\: \: \therefore$ the points are $A(3, 4, 2) , B(2, -2, -1), C(7, 0, 1)$
Step 2
The vector equation is of the form $ \overrightarrow r = (1-s-t)\overrightarrow a+s\overrightarrow b+t\overrightarrow c$
i.e., $ \overrightarrow r=(1-s-t)(3\overrightarrow i+4\overrightarrow j+2\overrightarrow k)+s(2\overrightarrow i-2\overrightarrow j-\overrightarrow k)+t(7\overrightarrow i+\overrightarrow k)$
Step 3
The cartesian equation is $ \begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix}=0$
where $ (x_1, y_1, z_1)=(3, 4, 2), (x_2, y_2, z_2)=(2, -2, -1), (x_3, y_3, z_3)=(7, 0, 1)$
$ \begin{vmatrix} x-3 & y-4 & z-2 \\ 2-3 & -2-4 & -1-2 \\ 7-3 & 0-4 & 1-2 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} x-3 & y-4 & z-2 \\ -1 & -6 & -3 \\ 4 & -4 & -1 \end{vmatrix}=0$
$ (x-3)(6-12)-(y-4)(1+12)+(z-2)(4+24)=0$
$ -6x+18-13y+52+28z-56=0 \Rightarrow 6x+13y-28z=14$

 

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

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