Question:
The equation of a plane containing the line of intersection of the planes \(2x-y-4=0\) and \(y+2z-4=0\), and passing through the point \((1,1,0)\), is
The equation of a plane containing the line of intersection of the planes \(2x-y-4=0\) and \(y+2z-4=0\), and passing through the point \((1,1,0)\), is
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A plane through the intersection of \(P_1=0\) and \(P_2=0\) is \(P_1+\lambda P_2=0\).
Updated On: Jul 3, 2026
- \(x-3y-2z=-2\)
- \(2x-z=2\)
- \(x-y-z=0\)
- \(x+3y+z=4\)
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The Correct Option is C
Solution and Explanation
Concept:
The equation of a plane passing through the line of intersection of two planes \[ P_1=0 \] and \[ P_2=0 \] is given by \[ P_1+\lambda P_2=0 \] where \(\lambda\) is a parameter.
Step 1: Write the given planes. \[ P_1=2x-y-4=0 \] \[ P_2=y+2z-4=0 \] Therefore, the required plane is \[ (2x-y-4)+\lambda(y+2z-4)=0 \]
Step 2: Use the given point \((1,1,0)\). Substitute \[ x=1,\quad y=1,\quad z=0 \] \[ (2(1)-1-4)+\lambda(1+2(0)-4)=0 \] \[ (2-1-4)+\lambda(1-4)=0 \] \[ -3-3\lambda=0 \] \[ \lambda=-1 \]
Step 3: Substitute \(\lambda=-1\). \[ (2x-y-4)-(y+2z-4)=0 \] \[ 2x-y-4-y-2z+4=0 \] \[ 2x-2y-2z=0 \] Divide by \(2\): \[ x-y-z=0 \]
Step 4: Final answer. \[ \boxed{x-y-z=0} \]
The equation of a plane passing through the line of intersection of two planes \[ P_1=0 \] and \[ P_2=0 \] is given by \[ P_1+\lambda P_2=0 \] where \(\lambda\) is a parameter.
Step 1: Write the given planes. \[ P_1=2x-y-4=0 \] \[ P_2=y+2z-4=0 \] Therefore, the required plane is \[ (2x-y-4)+\lambda(y+2z-4)=0 \]
Step 2: Use the given point \((1,1,0)\). Substitute \[ x=1,\quad y=1,\quad z=0 \] \[ (2(1)-1-4)+\lambda(1+2(0)-4)=0 \] \[ (2-1-4)+\lambda(1-4)=0 \] \[ -3-3\lambda=0 \] \[ \lambda=-1 \]
Step 3: Substitute \(\lambda=-1\). \[ (2x-y-4)-(y+2z-4)=0 \] \[ 2x-y-4-y-2z+4=0 \] \[ 2x-2y-2z=0 \] Divide by \(2\): \[ x-y-z=0 \]
Step 4: Final answer. \[ \boxed{x-y-z=0} \]
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