[member910132]

The perpendicular unit vectors i and j are in the horizontal plane and the unit vector k is directed vertically upwards.

Write down the first two values of t for which Tom is directly below his starting
point.

Here is what I did:
The i and j components must be 0, so







Now I cant get a value of t in which both are 0, let alone a value for which bot are 0 and the K component is negative.

Help is highly appreciated

Edit: The MS says that the answers are t = 2pi and t= 4 pi. But putting t=2pi in gives r = 2i + 3.7k, which can't be directly below his starting position as it has moved 2 units east and 3.7 units up ?? And the same problem occurs when t=4 is put in.

[a.partridge]

hint: when t = 0

i = 2
j = 0

[member910132]

(Original post by a.partridge)
hint: when t = 0

i = 2
j = 0

But I need i = 0, j = 0 and k to be negative.

[a.partridge]

(Original post by member910132)
But I need i = 0, j = 0 and k to be negative.

well no value exists where cos(t) and sin(t) are both 0

but that doesn't matter because at the starting point (i) and (j) are not both 0.

the starting point comes from r(0) = ........ you see not the origin!

[a.partridge]

regarding the negative you can see that the solutions must be the first to satisfy the inequality

(A)t > 10 there A is the coefficient in the expression for the displacement parallel to K

[member910132]

(Original post by a.partridge)
well no value exists where cos(t) and sin(t) are both 0

but that doesn't matter because at the starting point (i) and (j) are not both 0.

the starting point comes from r(0) = ........ you see not the origin!

Thnx a lot for that