Question:
The equation of the plane containing the line \( \frac{x-\alpha}{l} = \frac{y-\beta}{m} = \frac{z-\gamma}{n} \) is \( a(x-\alpha)+b(y-\beta)+c(z-\gamma)=0 \), where \( al + bm + cn \) is equal to
The equation of the plane containing the line \( \frac{x-\alpha}{l} = \frac{y-\beta}{m} = \frac{z-\gamma}{n} \) is \( a(x-\alpha)+b(y-\beta)+c(z-\gamma)=0 \), where \( al + bm + cn \) is equal to
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If a line lies in a plane, its direction ratios satisfy plane equation.
Updated On: Apr 30, 2026
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Solution and Explanation
Concept:
Direction ratios of the line must satisfy plane equation.
Step 1: Direction ratios of line. \[ (l, m, n) \]
Step 2: Plane contains line. Thus every point on line satisfies: \[ a(x-\alpha)+b(y-\beta)+c(z-\gamma)=0 \]
Step 3: Substitute parametric form. \[ x=\alpha+lt, y=\beta+mt, z=\gamma+nt \] Substitute: \[ a(lt)+b(mt)+c(nt)=0 \] \[ t(al+bm+cn)=0 \]
Step 4: Conclusion. Since true for all \(t\): \[ al+bm+cn=0 \]
Step 1: Direction ratios of line. \[ (l, m, n) \]
Step 2: Plane contains line. Thus every point on line satisfies: \[ a(x-\alpha)+b(y-\beta)+c(z-\gamma)=0 \]
Step 3: Substitute parametric form. \[ x=\alpha+lt, y=\beta+mt, z=\gamma+nt \] Substitute: \[ a(lt)+b(mt)+c(nt)=0 \] \[ t(al+bm+cn)=0 \]
Step 4: Conclusion. Since true for all \(t\): \[ al+bm+cn=0 \]
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