A) The nonzero vector orthogonal to the plane through the points P,Q and R is 0i -24j - 12k.
B) Area = 72√5
A) We are given the points P(2, 0, 2), Q(−2, 1, 4), R(6, 2, 6).
PQ = Q - P
PQ = (-2 - 2)i + (1 - 0)j + (4 - 2)k
PQ = -4i + j + 2k
PR = R - P
PR = (6 - 2)i + (2 - 0)j + (6 - 2)k
PR = 4i + 2j + 4k
The orthogonal vector is PQ * QR =
[tex]\left[\begin{array}{ccc}i&j&k\\-4&1&2\\4&2&4\end{array}\right][/tex]
= (4 - 4)i + (-16 - 8)j + (-8 - 4)k
= 0i -24j - 12k
Therefore the nonzero vector orthogonal to the plane through the points P,Q and R is 0i -24j - 12k
b) The area of the triangle with the vertices P,Q and R is the half of the length of the cross product of PQ and PR.
Thus;
Area = ¹/₂(PQ * PR)
Area = ¹/₂√((-24)² + (-12)²)
Area = ¹/₂√720
Area = 72√5
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