Answer:
a. <2,3,1>
b. [tex]\frac{\sqrt{14} }{2}[/tex]
Step-by-step explanation:
From the data given, we have point
P(1,3,5)
Q(-4,7,3)
R(2,2,6)
a. For us to find a vector orthogonal to the plane through the points P,Q and R we define two vectors such as
PQ=<-4,7,3>-<1,3,5>
PQ=<-5,4,-2>
we also define PR
PR=<2,2,6>-<1,3,5>
PR=<1,-1,1>
The cross product of the two formed vector gives rise to a vector that is orthogonal to the given plane Hence we carry out the cross product
[tex]\left[\begin{array}{ccc}i&j&k\\-5&4&-2\\1&-1&1\end{array}\right] \\i(4-2)-j(-5+2)+k(5-4)\\2i+3j+k\\[/tex]
Hence the required vector is <2,3,1>
b. The area of the formed triangle can be expressed as
[tex]area=\frac{1}{2}|PQxPR|\\but PQxPR=<2,3,1>\\ area=\frac{1}{2}\sqrt{(2)^{2}+(3)^{2}+(1)^{2}}\\ area=\frac{\sqrt{14}}{2}[/tex]
