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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the vector equation of the plane, passing through the points A(2,2,-1), B(3,4,2) and C(7,0,6). Also, find the cartesian equation of the plane.

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  • Cartesian form of the plane containing three non collinear points $(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ are $\begin{vmatrix}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\end{vmatrix}=0$
Step 1:
The given points are $A(2,2,-1),B(3,4,2)$ and $C(7,0,6)$
Let $\overrightarrow a=2\hat i+2\hat j-\hat k,\overrightarrow b=3\hat i+4\hat j+2\hat k$ and $\overrightarrow c=7\hat i+6\hat k$
Hence the vector equation of the line passing through the points.
$(\overrightarrow r-\overrightarrow a).(\overrightarrow{AB}\times \overrightarrow{AC})=0$
$(\overrightarrow r-\overrightarrow a).(\overrightarrow{b}-\overrightarrow{a}\times (\overrightarrow c-\overrightarrow{a})=0$
$\overrightarrow b-\overrightarrow a=(3\hat i+4\hat j+2\hat k)-(2\hat i+2\hat j-\hat k)$
$\Rightarrow \hat i+2\hat j+3\hat k$
$\overrightarrow c-\overrightarrow a=(7\hat i+6\hat k)-(2\hat i+2\hat j-\hat k)$
$\Rightarrow 5\hat i-2\hat j+7\hat k$
The required equation of the plane is $\overrightarrow r-(2\hat i+2\hat j-\hat k).[(\hat i+2\hat j+3\hat k)\times (5\hat i-2\hat j+7\hat k)=0$
Step 2:
$\begin{vmatrix}\hat i&\hat j&\hat k\\1 & 2 &3\\5 &-2&7\end{vmatrix}$
$\Rightarrow \hat i(14+6)-\hat j(7-15)+\hat k(-2-10)$
$\Rightarrow 20\hat i+8\hat j-12\hat k$
$\Rightarrow (\overrightarrow r-2\hat i+2\hat j-\hat k).20\hat i+8\hat j-12\hat k)=0$
This is required vector equation of the plane.
Step 3:
The cartesian equation is
$\begin{vmatrix}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\x_3-x_1&y_3-y_1&z_3-z_1\end{vmatrix}=0$
(i.e) $\begin{vmatrix}x-2&y-2&z+1\\3-2&4-2&2+1\\7-2&0-2&6+1\end{vmatrix}=0$
$(x-2)[14+6]-(y-2)[7-15]+(z+1)[-2-10]=0$
$\Rightarrow (x-2)(20)-(y-2)(-8)+(z+1)(-12)=0$
$\Rightarrow 20x-40+8y-16-12z-12=0$
$20x+8y-12z-68=0$
$5x+2y-3z-17=0$
This is required equation of the plane.
answered Nov 11, 2013 by sreemathi.v
 

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