Three coinitial vectors of magnitudes a, 2a and 3a meet at a point and their directions are along the diagonals if three adjacent faces if a cube. Determined their resultant R. Also prove that the sum of the three vectors determinate by the diagonals of three adjacent faces of a cube passing through the same corner, the vectors being directed from the corner, is twice the vector determined by the diagonal of the cube.

Explanation

Let the length of an edge of the cube be taken as unity and the vectors represented by OA, OB and OC (let the three coterminous edges of unit be `hati, hatj and hatk`, respectively). OR, OS and OT are the three diagonals of the three adjacent faces of the cube along which act the forces of magnitudes `a, 2a and 3a`, respectively. To find the vectors representing these forces, we will first find unit vectors in these directions and then multiply them by the corresponding given magnitudes of these forces.
Since `vec(OR) = hatj+hatk`, the unit vector along `OR` is `(1)/(sqrt(2))(hatj+hatk)`.

Hence, force `vec(F_(1))` of magnitude `a` along `OR` is given by
`" "vec(F_(1))=(a)/(sqrt(2))(hati+hatk)`
Similarly, force `vec(F_(1))` of magnitude 2a along OS is `(2a)/(sqrt(2))(hatk+hati)` and force `vec(F_(3))` of magnitude 3a along OT is `(3a)/(sqrt(2))(hati+hatj)`.
If `vecR` is their resultant, then
`" "vecR=vec(F_(1))+vec(F_(2))+vec(F_(3))`
`" " = (a)/(sqrt2) (hatj +hatk) + (2a)/(sqrt2) (hatk + hati) + (3a)/(sqrt2) (hati + hatj)`
`" " = (5a)/(sqrt2) hati + (4a)/(sqrt2) hatj + (3a)/(sqrt 2) hatk`
Again, `vec(OR) + vec(OS) + vec(OT) = hatj +hatk +hati +hatk +hati +hatj`
`" " = 2(hati +hatj +hatk)`
Also, `vec(OP) = vec(OT) + vec(TP) = (hati +hatj + hatk)" "(because vec(OT) = hati +hatj and vec(TP) = vec(OC) = hatk)`
`" " vec(OR) + vec(OS) + vec(OT) = 2 vec(OP) `