Three forces acting at a point are F1=2i-j+3k, F2=-i+3j+2k, F3=-i+2j-k.Find the directions and magnitude of F1+F2+F3, F1-F2+F3.
In order to add 3 vectors, F1, F2, F3, we'll add or subtract algebraically the coefficients of correspondent unit vectors: i,j,k.
F1+F2+F3 = (2i-j+3k) + (-i+3j+2k) + (-i+2j-k)
We'll remove the brackets and combine like terms:
F1+F2+F3 = i(2-1-1) + j(-1+3+2) + k(3+2-1)
F1+F2+F3 = 0i + 4j + 4k
So, the resultant vector of the sum of 3 vectors F1+F2+F3 has no component in the x direction, but it has a component of 4 units in y direction and a component of 4 units in z direction.
The magnitude of the resultant vector is:
|F1+F2+F3| = sqrt (0^2 + 4^2 + 4^2)
|F1+F2+F3| = sqrt 32
|F1+F2+F3| = 4sqrt2
|F1+F2+F3| = 5.66 units
The resultant vector has a magnitude of 5.66 units and it is located in y-z plane. The vector makes an angle with y axis.
F2 = -i+3j+2k
F3 = -i+2j-k.
Therefore F1+F2+F3 = (2i-j+3k)+(-i+3j+2k)+(-i+2j-k)= (2-1-1)i+(-1+3+2)j+(3+2-1)k = 0i+4j+4k.
Magnitude of the vector sum =|F1+F2+F3| = |0+4j+4k| = sqrt32 = 4sqrt2
Direction of F1+F2+F3 = (0/4sqrt2 . 4/4sqrt2, 4/4sqrt2) = (0, 1/sqr2, 1/sqrt2)
F1-F2+F3 = (2i-j+3k)-(-i+3j+2k)+(-i+2j-k)= (2+1-1)i+(-1-3+2)j+(3-2-1)k = 2i-2j+0k
Threfore manitude of F1-F2+F3 = |F1-F2+F3| = |2i-2j+0k| = sqrt(2^2+2^2) = 2sqrt2 .
Direction of F1-F2+F3 = (2/2sqrt2 , -2/2sqrt2 , 0/2sqrt2) = 1/sqrt2, -1/sqrt2 , 0).