Find the area of the parallelogram with adjacent sides u = –4i – 9j + k and v = –6i + j + 5k. please just give the answer.

75 (approx.)

Step-by-step 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.