Question

If force F1 = 2i+2j and f2=3i+4k are acting on a particle component of F1 perpendicular to F2 is?

Answer 1

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.

NEELA | STUDENT

F1 =2i-j+3k

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).