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Question
Reduce the equations of the following planes to intercept form and find the intercepts on the coordinate axes.
4x + 3y − 6z − 12 = 0
Solution
Equation of the given plane is
\[4x + 3y - 6z - 12 = 0\]
\[ \Rightarrow 4x + 3y - 6z = 12\]
\[\text{ Dividng both sides by 12, we get } \]
\[ \frac{4x}{12} + \frac{3y}{12} + \frac{( - 6z)}{12} = \frac{12}{12}\]
\[ \Rightarrow \frac{4x}{12} + \frac{3y}{12} - \frac{6z}{12} = \frac{12}{12}\]
\[ \Rightarrow \frac{x}{3} + \frac{y}{4} + \frac{z}{- 2} = 1 . . . \left( 1 \right)\]
\[\text{ We know that the equation of the plane whose intercepts on the coordianate axes are a,b and c is } \]
\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 . . . \left( 2 \right)\]
\[\text{ Comparing (1) and (2), we get }\]
\[a = 3; b = 4; c = - 2\]
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