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Home  >>  TN XII Math  >>  Vector Algebra
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Derive the equation of the plane in the intercept form.

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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)
Step 1
Let the plane meet the coordinate axes at A, B, C where $OA=a, \: OB = b,\: OC=C$ ( the intercepts). $ \therefore $ the plane passes through the points $A(a, 0, 0), B(0,  b,  0 ), C(0, 0, c).$
Step 2
The cartesian equation of the plane 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)=(a, 0, 0)(x_2, y_2, z_2)=(0, b, 0) , (x_3, y_3, z_3)=(0, 0, c)$
$ \therefore \begin{vmatrix} x-a & y & z \\ -a & b & 0 \\ -a & 0 & c \end{vmatrix}=0$
$ (x-a)(bc-0)-y(-ac-0)+z(0+ab)=0$
$xbc-abc+yac+zab=0$ ( divided by abc)
$ \large\frac{x}{a} +\large\frac{y}{b}+\large\frac{z}{c}=1$. This is the intercepts form of equation of the plane.

 

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

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