Question

How do you find a unit vector that is orthogonal to both $u = \left( {1,0,1} \right)$ $v = \left( {0,1,1} \right)$?

Answer 1

In the given question, we have been given two vectors. In this question, we have to find a third vector which is both a unit vector and orthogonal to the two given vectors. To do that, we are going to first represent the ordered triplets of the given vectors as some variables. Then we are going to put them in a particular formula and evaluate the value for the vector to be orthogonal. Then we have to divide the obtained orthogonal vector by its magnitude so as to make it unit too.

Complete step by step solution:
The given two vectors are $u = \left( {1,0,1} \right)$ and $v = \left( {0,1,1} \right)$.
First, we find the cross product of the two vectors.
$u \times \;v = \left( {1,0,1} \right) \times \left( {0,1,1} \right)$
Applying the formula of cross product and solving,
$= \left( {\left| \begin{array}{l}0{\rm{ }}1\\\1{\rm{ }}1\end{array} \right|,\left| \begin{array}{l}1{\rm{ }}1\\\1{\rm{ }}0\end{array} \right|,\left| \begin{array}{l}1{\rm{ }}0\\\0{\rm{ }}1\end{array} \right|} \right) = \left( { - 1, - 1,1} \right)$
Now, $\left| {\left( { - 1, - 1,1} \right)} \right| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt 3$
Hence, to make $\left( { - 1, - 1,1} \right)$ into a unit vector, we divide it by $\sqrt 3$,
$\left( {\dfrac{{ - 1}}{{\sqrt 3 }},\dfrac{{ - 1}}{{\sqrt 3 }},\dfrac{1}{{\sqrt 3 }}} \right)$

Note: In the given question, we had been given two vectors. We had to find a third vector which is both a unit vector and orthogonal to the two given vectors. We solved it by putting them in a particular formula and evaluating the value for the vector to be orthogonal. So, it is very important that we know all the formulae of the given question’s type, how to use them in the given scenario and how to transform into the given form.