The three vertices of a parallelogram PQRS are P(3,-1,2), Q(1,2,-4), R(-1,1,2). Then, the coordinates of the fourth vertex is .
A
(-1,-2,8)
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B
(1, 2,8)
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C
(1,-2,8)
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D
(1,-2,-8)
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Solution
The correct option is C (1,-2,8) Let the coordinates of the fourth vertex S be (x,y,z). Since, the diagonals of a parallelogram bisect each other. So,the mid-points of PR and QS will coincide. ∴(3−12,−1+12,2+22)=(1+x2,2+y2,−4+z2)⇒(1,0,2)=(x+12,y+22,z−42)⇒x+12=1,y+22=0andz−42=2⇒x=1,y=−2andz=8