perpendicular vectors

[M-D]

hi,

i have a vector with coordinates: (0,3,4). two unit vectors which are perpendicular to this vector are: (0,-1/3,1/4) and (0,1/3,-1/4). are these the only two unit vectors perpendicular to (0,3,4) or are there more?

thanks in advance.

[Limista]

There are many more.

You have a vector   a = 3j + 4k

You can find any perpendicular vector b as long as a.b = 0 according to dot product rule, where b = xi + yj + zk

[M-D]

thank you for your reply. for me to be able to solve for x,y and z (3 unknowns) i must have 3 equations. here i only have 1, so how can i solve for x, y and z?

thanks

[Limista]

Actually you only have 2 unknowns, which are y and z.

When you do a.b

(0*x)i + 3yj + 4zk = 0

3yj + 4zk = 0

Substitute in any value you would like for y, and then find z.

Or substitute in any value you would like for z, and then find y.

As part of your perpendicular vector, you can also let x be any number at all, because it won't matter (multiplying by 0 anyway as demonstrated above). You don't even have to have an x actually.

[M-D]

thank you very much. i appreciate your help. i wanted to ask one other thing. the two vectors i initially found were  (0,-1/3,1/4) and (0,1/3,-1/4). in the question it says that these must be unit vectors but i don't think the ones i have found really have a length of 1 (is that correct?) which is what a unit vector essentially is. how should i go about converting them to unit vectors if they are not unit vectors?

[brightsky]

divide by magnitude

[nubs]

If I have a vector, a = xi + yj + zk
Then the unit vector is 1 divided by the magnitude of a multiplied by the vector a

So if I had 2i + 3j +1k
The magnitude would be (4+9+1)^(1/2)
The unit vector would then be 1 divided by that magnitude, multiplied by (2i + 3j +1k)

[M-D]

yes i understand. thanks