Question:
Find equation of plane passing through (1,2,3), (-1,4,2), (3,1,1)
Find equation of plane passing through (1,2,3), (-1,4,2), (3,1,1)
Show Hint
Cross product gives direction perpendicular to plane.
Updated On: May 1, 2026
- \(5x+y+12z-23=0\)
- \(5x+6y+2z=23\)
- \(x+6y+2z=13\)
- \(x+y+z=13\)
- \(2x+6y+5z=7\)
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The Correct Option is C
Solution and Explanation
Concept: Use cross product to find normal vector.
Step 1: Form vectors: \[ \vec{AB}=(-2,2,-1),\quad \vec{AC}=(2,-1,-2) \]
Step 2: Compute cross product: \[ \vec{n}=\vec{AB}\times \vec{AC} \]
Step 3: Determinant: \[ \begin{vmatrix} i & j & k -2 & 2 & -1 2 & -1 & -2 \end{vmatrix} \]
Step 4: Expand: \[ i(-4-1) - j(4+2) + k(2-4) = (-5,-6,-2) \]
Step 5: Plane: \[ -5(x-1)-6(y-2)-2(z-3)=0 \]
Step 6: Simplify: \[ 5x+6y+2z=13 \]
Step 1: Form vectors: \[ \vec{AB}=(-2,2,-1),\quad \vec{AC}=(2,-1,-2) \]
Step 2: Compute cross product: \[ \vec{n}=\vec{AB}\times \vec{AC} \]
Step 3: Determinant: \[ \begin{vmatrix} i & j & k -2 & 2 & -1 2 & -1 & -2 \end{vmatrix} \]
Step 4: Expand: \[ i(-4-1) - j(4+2) + k(2-4) = (-5,-6,-2) \]
Step 5: Plane: \[ -5(x-1)-6(y-2)-2(z-3)=0 \]
Step 6: Simplify: \[ 5x+6y+2z=13 \]
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