A plane passes through the points ((1,0,2),;(2,3,0),;(4,-1,1)). Compute the perpendicular distance from the point ((5,-1,2)) to this plane.
Answer
(\frac{sqrt{6}}{2})
The correct answer is (\frac{sqrt{6}}{2}) — Rationalize if desired: (\frac{3sqrt{6}}{6}=\frac{sqrt{6}}{2}).
Step-by-step solution
- Find two direction vectors in the plane: (mathbf{a}=langle1,3,-2\rangle) (from ((1,0,2)) to ((2,3,0))) and (mathbf{b}=langle3,-1,-1\rangle) (from ((1,0,2)) to ((4,-1,1))).
- Compute normal vector: (mathbf{n}=mathbf{a}\times\mathbf{b}=langle-5,-5,-10\rangle).
- Simplify to (langle1,1,2\rangle) after dividing by (-5).
- Equation of plane: (1(x-1)+1(y-0)+2(z-2)=0) → (x+y+2z-5=0).
- Plug point ((5,-1,2)) into distance formula: (\frac{|5+(-1)+4-5|}{sqrt{1^2+1^2+2^2}}=\frac{|3|}{sqrt{6}}=\frac{3}{sqrt{6}}).
- Rationalize if desired: (\frac{3sqrt{6}}{6}=\frac{sqrt{6}}{2}).
Common mistake: Using a wrong normal vector or incorrect sign in the plane equation
Concept tested: Distance from point to a plane
Published · Last updated · How this solution was prepared and checked
← All Three Dimensional Geometry questions