The area of parallelogram with the vertices is √269 when vertices are P(1, 0, 2), Q(3, 3, 3), R(7, 5, 8), and S(5 , 2, 7).
Given that,
The vertices are P(1, 0, 2), Q(3, 3, 3), R(7, 5, 8), and S(5 , 2, 7).
We have to find the area of parallelogram with the vertices.
We know that,
PQ=Q-P=(3,3,3) -(1,0,2)=(2,3,1)
PR=R-P=(7,5,8) -(1,0,2)=(6,5,6)
Find area using cross product as
Area= |PQ x PR|
=|i (3(6)-1(5)) - j (2(6)-1(6)) + k (2(5)-3(6)) |
= |i (18 - 5) - j (12 -6) + k (10 -18) |
= |{13; 6; -8}|
=√(13²+6²+(-8)²)
=√269
Therefore, the area of parallelogram with the vertices is √269 when vertices are P(1, 0, 2), Q(3, 3, 3), R(7, 5, 8), and S(5 , 2, 7).
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