# Vector linear relationship

1. Apr 29, 2017

### Taylor_1989

1. The problem statement, all variables and given/known data
Find the general solution to the equation: ${u}\times{(i+4j)}=3k$
2. Relevant equations
$u=(ai+bj+ck)$

3. The attempt at a solution
I am having trouble visualising what to do here.

So my throught so far is that, if i do $(ai+bj+ck)\times {(i+4j)}=(0i-0j+(4a-b)k)$
now if I equate $(0i-0j+(4a-b)k)=(0i+0j+3k)$ I get $4a-b=3$ and this is where i draw a blank,it's the general equation part that I dont get.

I mean if I rearange I could do for a or b, I preffer b as it have no fractions, $4a-3=b$, so I assume that I would sub this back into u which would give me $u=(0i+(4a-3)j+0k)$. Now doe the general equation, mean vector equation of the line. Normally I would find the direction vector and the starting vector but it being a cross product has thrown me a bit. Am I on the right lines or have I gone very wrong?

2. Apr 29, 2017

### Staff: Mentor

From the first you know that u is perpendicular to k hence u can be written as ai + bj and from there sub into the cross product and solve it.

3. Apr 29, 2017

### Taylor_1989

Hi, is that not what I have done, the ans in the back of the exercsie book gives the ans as: $u=-3j+t(i+4j)$ where has the parameter t come into, p.s I did not look at the ans before I posted, but still unawre how u can equal this, as to me it look like the vector equation of a line.

4. Apr 29, 2017

### Buffu

You just need to equate magnitudes. since any two dimensional vector have $k$ as the unit vector of vector product.

5. Apr 29, 2017

### Taylor_1989

Ok, I am really not understanding here, how dose equating the magnitudes help with this problem as there is a parameter t. Could you please expand, if you calculate the magnitude surely you just working out the area of a parallelogram?

6. Apr 29, 2017

### ehild

$\vec u=a \vec i + b \vec j +c \vec k$. You obtained b=4a-3 which is correct, and a is arbitrary. For the general solution, you can not take the coefficient of $\vec i$ zero in vector $\vec u$.

7. Apr 29, 2017

### Taylor_1989

so from what u are saying is that $a=t, u=ai+bj=ti+(4t-3)j$ now I assume that this is t can be any number along the u vector?

8. May 1, 2017

### ehild

t is an arbitrary number, and u depends on it. "number along the u vector" has no sense.

9. May 3, 2017

### Taylor_1989

Thank you for you help after taking a step back I relieased what was going on. Once again thank you for the help