Respuesta :
Answer:
a. (23,-20,-5)
b. A=15.4
Step-by-step explanation:
first we find two vectors on the plan with length p q and PR.
We have
P Q=<5,1,-3> - <0,-4,0>
P Q=<5-0,1-(-4),-3-0>
P Q=(5,5,-3)
PR=<5,2,1> - <0,-4,0>
PR=<5-0,2-(-4),1-0>
PR=(5,6,1)
The cross product of the two determined vector will produce an orthogonal to the plane
the cross product of two vectors is expressed as
(a,b,c) x (d,e,f)=(bf-c e, c d-a f, a e-b d)
going by the expansion above we have
P Q * PR =(5,5,-3) X (5,6,1) =(5-(-18), -15-5, 30-25)
P Q * PR =(23,-20,-5).
b. To determine the area, we use also use the cross product of the two formed vectors i.e
[tex]A=\frac{1}{2}|PQxPR|\\[/tex]
[tex]A=\frac{1}{2}|(5,5,-3)x(5,6,1)|\\A=\frac{1}{2}|(23,-20,-5)|\\\\A=\frac{1}{2}\sqrt{23^{2}+20^{2}+5^{2}} \\A=\frac{1}{2}(30.9)\\A=15.4 unit square[/tex]