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.