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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
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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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