Question

Find two unit vectors u1 and u2 that are perpendicular to a=i+2j-k and b=i+j+2k

Answer 1

When 3d vectors are involved, a fail-safe way to obtain a vector that is perpendicular to both of them is via the cross-product :)

Answer 2

And in this case you'll have the positive and negative ones, though I'm not sure if that counts as two unit vectors... I think it would.

Answer 3

To get the cross-product of two vectors \(\large \vec{v_1}\) and \(\large \vec{v_2}\) where
\[\Large \vec{v_1}= a\vec{i}+b\vec{j}+c\vec{k}\]\[\Large \vec{v_2}= p\vec{i}+q\vec{j}+ r\vec{k}\]

Use this determinant
\[\LARGE \left|\begin{matrix} \vec{i}&\vec{j}&\vec{k}\\a&b&c\\p&q&r\end{matrix}\right|\]

Answer 4

\[\huge \vec{v_1}\times \vec{v_2} =\left|\begin{matrix} \vec{i}&\vec{j}&\vec{k}\\a&b&c\\p&q&r\end{matrix}\right|\]

Answer 5

Which equals \[ i^\rightarrow(br-qc)+ j^\rightarrow(cp-ar)+k^\rightarrow(aq-bp)\] in case determinants are unfamiliar. Basically, diagonals which go down and right are positive, diagonals which go down and left are negative, and verticals are zero.

Answer 6

@Numb3r1 :)
for future reference... \vec{} puts a right-facing arrow on top of whatever is inside those braces...

@Chelsea04
So that'd give you a vector, the other vector would be
\[\huge \vec{v_2}\times \vec{v_1}\]
but don't bother with this if you've already gotten the first cross-product
\[\huge \vec{v_2}\times \vec{v_1}= -(\vec{v_1}\times \vec{v_2})\]
so just get the negatives...


I'd like to emphasise that these two vectors you get may well not be unit vectors, you have to divide them by their magnitude.

And you'll be done :)

Answer 7

I was wondering about that! @terenzreignz