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Reduce the Equations of the Following Planes to Intercept Form and Find the Intercepts on the Coordinate Axes. (Ii) 2x + 3y − Z = 6 - Mathematics

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प्रश्न

Reduce the equations of the following planes to intercept form and find the intercepts on the coordinate axes. 

 2x + 3y − z = 6

योग

उत्तर

The equation of the given plane is

\[ 2x + 3y - z = 6\]

\[ \text{ Dividng both sides by 6, we get } \]

\[ \frac{2x}{6} + \frac{3y}{6} - \frac{z}{6} = \frac{6}{6}\]

\[ \Rightarrow \frac{x}{3} + \frac{y}{2} + \frac{z}{- 6} = 1 . . . \left( 1 \right)\]

\[\text{ We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c  is } \]

\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 . . . \left( 2 \right)\]

\[\text{ Comparing (1) and (2), we get } \]

\[a = 3; b = 2; c = - 6\]

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अध्याय 29: The Plane - Exercise 29.02 [पृष्ठ ७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 29 The Plane
Exercise 29.02 | Q 2.2 | पृष्ठ ७

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