Answer:
75 (approx.)
Explanation:
Cross Products
Cross product of two non-collinear vectors is a third vector which is orthogonal to the first two. The magnitude of the cross product equals the area of a parallelogram having the two given vectors as sides.
The the cross product u x v of vectors u<u1,u2,u3> and v<v1,v2,v3> is the value of the determinant formed by the direction vector <i,j,k> and u, v :
i j k
u1 u2 u3
v1 v2 v3
which evaluates to
P = u x v = (u2*3 - u3*v2)i – (u1*v3 - v1*u3)j + (u1*v2 – v1*u2)k = <p1, p2, p3>
The magnitude, which equals the area of the parallelogram, is the norm of P
Area = ||P|| = sqrt(p1²+p2² +p3²)
Plugging in values,
u<-4,-9,1>
v<-6,1,5>
Cross product P=(-45-1)i+(20-6)j+(-54-4)k =<-46+14-58>
Magnitude = sqrt((-46)^2+(14)^2+(-58)^2) = 2sqrt(1419)=75.34 (approx.)
So the area of the parallelogram is 75.3 approx.