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