Task: Question Answering
Input (English): Passage: Plane (geometry) -- The remainder of the expression is arrived at by finding an arbitrary point on the line. To do so, consider that any point in space may be written as r = c 1 n 1 + c 2 n 2 + λ ( n 1 × n 2 ) (\displaystyle (\mathbf (r))=c_(1)(\mathbf (n))_(1)+c_(2)(\mathbf (n))_(2)+\lambda ((\mathbf (n))_(1)\times (\mathbf (n))_(2))) , since ( n 1 , n 2 , ( n 1 × n 2 ) ) (\displaystyle \((\mathbf (n))_(1),(\mathbf (n))_(2),((\mathbf (n))_(1)\times (\mathbf (n))_(2))\)) is a basis. We wish to find a point which is on both planes (i.e. on their intersection), so insert this equation into each of the equations of the planes to get two simultaneous equations which can be solved for c 1 (\displaystyle c_(1)) and c 2 (\displaystyle c_(2)) .
Question: can one point be in two different planes
Output (English): 1