# Thread: General solution of vector equation

1. ## General solution of vector equation

Find the general solution of the vector equation:
u $\times$(i-3k)=2j

I've got one more question similar, so i need some pointers.
Thanks!

2. Originally Posted by arze
Find the general solution of the vector equation:
u $\times$(i-3k)=2j

I've got one more question similar, so i need some pointers.
Thanks!
Substitute u = ai + bj + ck. Note that u.(2j) = 0 and so b = 0. So u = ai + ck. Substitute and solve for a and c. (Note that the cross-product gives one equation and you can get another equation by equating magnitudes).

3. I don't understand how u.(2j) = 0.

4. Ok got the part about b=0.
I found that 3a+c=2
I don't know what else.
The answer is u=2k+t(i-3k) where t is a scalar, i think

5. Originally Posted by arze
Ok got the part about b=0.
I found that 3a+c=2
I don't know what else.
The answer is u=2k+t(i-3k) where t is a scalar, i think
Calculate (ai + ck) x (i - 3k) and equate the result to 2j.

6. I found 3a+c=2, i need another equation of a and c?

7. ux(i-3k) is a vector- it has i, j, and k components. Setting the i component equal to 0, the j component equal to 2 and the k component equal to 0 gives you three equations for the three coefficients of i, j, and k in u.

8. Originally Posted by arze
I found 3a+c=2, i need another equation of a and c?
Go back and read post #2 more carefully.

9. Really sorry, but I can't get it, you said from the magnitudes? How do I find that?
Sorry I don't understand
Thanks

10. Originally Posted by arze
Really sorry, but I can't get it, you said from the magnitudes? How do I find that?
Sorry I don't understand
Thanks
If the answer has a parameter (t) in it, then you don't need another equation. Just get c in terms of a, say, and substitute into u = ai + ck.

Otherwise, (and I haven't checked whether the equation is different) equate the magnitude of each side of (ai + ck)(i-3k)=2j (I assume you know how to find the magnitude of a vector).