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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 points $(1 , 2 , 3 )$and $(2 , 3 , 1 )$ perpendicular to the plane $ 3x-2y+4z-5=0$

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  • Equation of a plane through points $A(x_1, y_1, z_1)(pv \overrightarrow a)$ and $B(x_2, y_2, z_2)(pv\: \overrightarrow b)$ and parallel to a vector $\overrightarrow u$ (with D.r.s $l, m, n)$ $ \overrightarrow r = (1-s) \overrightarrow a+s\overrightarrow b+t\overrightarrow v$ (vector form) $ \begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l & m & n \end{vmatrix}=0 $ ( cartesian form) $ [ \overrightarrow r-\overrightarrow a\: \: \overrightarrow b-\overrightarrow a\: \: \overrightarrow u]=0$ ( Non parametre vector equation)
  • 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 points
$A(1, 2, 3) [\overrightarrow a=\overrightarrow i+2\overrightarrow j+3\overrightarrow k]\: and \: B(2, 3, 1)[\overrightarrow b=2\overrightarrow i+3\overrightarrow j+\overrightarrow k]$
It is perpendicular to the plane $3x-2y+4z-5=0$ and therefore parallel to the normal vector $ \overrightarrow v=3\overrightarrow i-2\overrightarrow j+4\overrightarrow k$
Step 2
The vector equation of the plane is of the form
$ \overrightarrow r = (1-s)\overrightarrow a+s\overrightarrow b+t\overrightarrow v$
$ \overrightarrow r=(1-s)(\overrightarrow i+2\overrightarrow j+3\overrightarrow k)+s(2\overrightarrow i+3\overrightarrow j+\overrightarrow k)+t(3\overrightarrow i-2\overrightarrow j+4\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 \\ l & m & n \end{vmatrix}=0$
where $(x_1, y_1, z_1)=(1, 2, 3) , (x_2, y_2, z_2)=(2, 3, 1), (l, m, n)=(3, -2, 4)$
$ \begin{vmatrix} x-1 & y-2 & z-3 \\ 2-1 & 3-2 & 1-3 \\ 3 & -2 & 4 \end{vmatrix}=0 $ i.e., $ \begin{vmatrix} x-1 & y-2 & z-3 \\ 1 & 1 & -2 \\ 3 & -2 & 4 \end{vmatrix}=0$
$(x-1)(4-4)-(y-2)(4+6)+(z-3)(-2-3)=0$
$ \Rightarrow -10y+20-5+15=0\: or \: 2y+z=7$
answered Jun 16, 2013 by thanvigandhi_1
 

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