For is mathematically given as
Considering line segments from (0,0,0)origin to (3,0,0) , (3,5,1) and (0,5,1) under the infulence of force
Generally, the equation for force is mathematically given as
F = z2i + 5xyj + 2y2k
Therefore, Considering u*v
[tex]u * v = (u_1j+u_2j+u_3k) * (v_1i+v_2j+v_3k)[/tex]
[tex]u* v = u_1v_1(i * i) + u_1v_2(i * j)+u_1v_3(i * k) + u_2v_1(j * i) + u_2v_2(j * j)+u_2_3(j* k) + u_3v_1(k * i) + u_3v_2(k * j)+u_3v_3(k * k)[/tex]
Where
[tex]i * i = j *j=k * k=0[/tex]
Hence
[tex]u* v = u_1v_2k-u_1v_3j-u_2v_1k+u_2v_3i +u_3v_1j - u_3v_2i[/tex]
[tex]u = (u_1,u_2,u_3) = (0,5,1)\\\\v = (v_1,v_2,v_3) = (-3,0,0)[/tex]
The normal equation formed
-3y + 15z = 0
z= (1/5)y
Considering the level surface and differential surface area
h(x,y,z) = -y + 5z =0
[tex]dS = |grad(h)| dA[/tex]
In terms of the x and y coordinates of (3,0,0) and (3,5,1) and (0,5,1), we can state that the ranges are 0 to 3 and 5 respectively.
In conclusion,
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