Answer
(a) $\lt 0,18,-9 \gt$
(b) $\frac{9}{2} \sqrt 5$
Work Step by Step
(a) Given: $P(1,0,1),Q(-2,1,3)$ and $R(4,2,5)$
$PQ ^\to=\lt -2-1,1-0,3-1\gt=\lt -3,1,2 \gt$
$PR ^\to=\lt4-1,2-0,5-1\gt=\lt 3,2,4 \gt$
$\lt -3,1,2 \gt \times \lt 3,2,4 \gt =\lt 1(4)-2(2),2(3)-4(-3),(-3)2-1(3)\gt =\lt 0,18,-9 \gt$
(b) Area of a vector with vertices at P,Q, and R is
$ Area=\frac{1}{2}|PQ ^\to \times PR ^ \to|$
$PQ ^\to \times PR ^ \to=\lt -3,1,2 \gt \times \lt 3,2,4 \gt =\lt 1(4)-2(2),2(3)-4(-3),(-3)2-1(3)\gt =\lt 0,18,-9 \gt$
$|PQ ^\to \times PR ^ \to|=\sqrt {0^2+(18)^2+(-9)^2}=9 \sqrt 5$
$ Area=\frac{1}{2}|PQ ^\to \times PR ^ \to|=\frac{9}{2} \sqrt 5$
