The nonzero vector orthogonal to the plane through the points P,Q and R is -7i + 7j -21k and the area of the triangle is 11.6 square units.
(a)Given points are P(-1,3,1) , Q(0,7,2) and R(4,2,-1)
PQ = Q - P
= (0 - (-1))i + (7 - 3)j + (2 -1)k
=i + 4j + k
PR = R - P
= (4 - (-1))i + (2 - 3)j + (-1 - 1)k
5i -j - 2k
The orthogonal vector is PQ x QR
[tex]\left[\begin{array}{ccc}i&j&k\\1&4&1\\5&-1&-2\end{array}\right][/tex]
(-8 + 1)i - (-2 - 5)j + (-1 - 20))k
-7i + 7j - 21k
Therefore the nonzero vector orthogonal to the plane through the points P,Q and R is -7i + 7j - 21k
(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 QR.
Area = 1/2| PQ x QR|
= 1/2 √(7² + (-7)² + (-21)²)
= 1/2 √(49 + 49 + 441)
= 1/2 (23.21)
= 11.60
= 11.60 square units.
Therefore The non zero vector is -7i + 7j - 21k and the area of the triangle is 11 .6 square units.
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