Thus there is a unique plane passing through p q and

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. Thus there is a unique plane passing through P, Q, and R , namely the graph of 5 x + 16 y 6 z = 1 , obtained by cancelling the common factor s . 005 10.0 points
Version 015 – EXAM01 – gilbert – (56780) 3 3. 1 + 7 s 2 3 s 1 s , 0 s 1 . 4. 7 3 s 3 3 s 1 s , 0 s 1 . 5. 4 + 4 s 1 6 s 2 2 s , 0 s 1 . 6. 5 s 1 5 s 3 5 s , 0 s 1 . Explanation: The face CDEG of the parallelepiped lies in the unique plane in which the vertices D, C, and G lie. Now in vector form this plane is x = c + s b + t a , −∞ ≤ s, t ≤ ∞ , and the set of all points in the parallelogram CDEG corresponds to c + s b + t a , 0 s, t 1 . 007 10.0 points 006 10.0 points Explanation: If { x , y , z } is a linearly dependent set, then c 1 x + c 2 y + c 3 z = 0 for some choice of c 1 , c 2 , c 3 , not all zero. Now if c 3 = 0, then c 1 x + c 2 y = 0 . But if x , y are linearly independent, then c 1 = c 2 = 0. Thus c 3 ̸ = 0, so z = ( c 1 /c 3 ) x ( c 2 /c 3 ) y , i.e. , z is a linear combination of x , y . Hence z is a vector in Span { x , y } when x , y are lin- early independent and { x , y , z } is a linearly dependent set. Consequently, the statement is TRUE .
If x and y are linearly independent but { x , y , z } is a linearly dependent set, then z is in Span { x , y } . True or False? 1. TRUE correct 2. FALSE 3. x 1 = 3 4. x 1 = 1 + t, t arbitrary 5. x 1 = 1 + t, t arbitrary 6. no solution x 1 exists correct Explanation:
Version 015 – EXAM01 – gilbert – (56780) 4 By row reduction B = 1 1 2 2 0 0 2 4 0 0 1 0 1 1 2 2 0 0 2 4 0 0 0 2 , which is now in echelon form. But the system x 1 x 2 + 2 x 3 = 2 , 2 x 3 = 4 , 0 x 3 = 0 , associated with this matrix is inconsistent be- cause there is no solution x 3 . Consequently, no solution x 1 exists because the original sys- tem is inconsistent . 008 10.0 points

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