Let OD = i + 2 j + 6 k , CB = -3 i - 2 k be the diagonals of the parallelogram OBDC and OA = i + 2 j + 3 k be another vector. Then the volume of a parallelopiped determined by vectors OA , OB , and OC (in cubic units), is

Step-by-step solution

Let OB = b and OC = c . In the parallelogram OBDC, the diagonals are given by the sum and difference of the adjacent sides: OD = b + c = i + 2 j + 6 k CB = b - c = -3 i - 2 k The volume of the parallelopiped determined by vectors OA = a , OB = b , and OC = c is given by the scalar triple product V = | a ( b c )| . Taking the cross product of the diagonals: ( b + c ) ( b - c ) = b b - b c + c b - c c Since b b = c c = 0 and c b = - b c , we get: ( b + c ) ( b - c ) = -2( b c ) Now, computing the cross product of OD