Question

Find a unit vector u that is orthogonal to a and b where

(see attached image)

Answer 1

vectors a & b attached.

i understand that for a vector to be orthogonal a.b = 0 where can i go from there?

Answer 2

@zepdrix

Answer 3

a vector \(\vec{C}\) that is orthogonal to both \(\vec A\) and \(\vec B \) is the cross product \(\vec A \times \vec B\)

Answer 4

so just find C and find it's unit vector, is that it?

Answer 5

that's right

Answer 6

hint:
we can write this:
\[{\mathbf{a}} \times {\mathbf{b}} = \left| {\begin{array}{*{20}{c}}
{{\mathbf{\hat i}}}&{{\mathbf{\hat j}}}&{{\mathbf{\hat k}}} \\
1&8&{ - 3} \\
1&3&7
\end{array}} \right| = {\mathbf{\hat i}}\left( {56 + 9} \right) - {\mathbf{\hat j}}\left( {7 + 3} \right) + {\mathbf{\hat k}}\left( {3 - 8} \right) = ...\]

Answer 7

please compute the lenght of the vector \({\mathbf{a}} \times {\mathbf{b}}\)

Answer 8

michele in a x b is k 5 or -5? i got mine as 5 but yours is 5

Answer 9

@Michele_Laino

Answer 10

i used the method of setting up i j k -i -j to compute the cross product

Answer 11

I got this:
\[\begin{gathered}
{\mathbf{a}} \times {\mathbf{b}} = \left| {\begin{array}{*{20}{c}}
{{\mathbf{\hat i}}}&{{\mathbf{\hat j}}}&{{\mathbf{\hat k}}} \\
1&8&{ - 3} \\
1&3&7
\end{array}} \right| = {\mathbf{\hat i}}\left( {56 + 9} \right) - {\mathbf{\hat j}}\left( {7 + 3} \right) + {\mathbf{\hat k}}\left( {3 - 8} \right) = \hfill \\
\hfill \\
= 65{\mathbf{\hat i}} - 10{\mathbf{\hat j}} - 5{\mathbf{\hat k}} \hfill \\
\end{gathered} \]

Answer 12

so the corresponding length is:
\[\left| {{\mathbf{a}} \times {\mathbf{b}}} \right| = \sqrt {{{65}^2} + {{10}^2} + {5^2}} = \sqrt {...} \]

Answer 13

this is how i was thought to do the cross product, so i get axb = i(45+9)+j(-3-7)+k(8-3)

i know that method to but one gives me -5k and this method give me 5k which one is correct?

Answer 14

I don't know, I think that my computation is correct

Answer 15

got it thanks!