# Determine unit vector which is perpendicular to both A=2i+j+k and B=i-j+2k?

Sep 15, 2016

We know that cross product of any two vectors yields a vector which is perpendicular to both vectors
$\therefore$ for two vectors $\vec{A} \mathmr{and} \vec{B}$ if $\vec{C}$ is the vector perpendicular to both.
$\vec{C} = \vec{A} \times \vec{B} =$$\left[\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ {A}_{1} & {A}_{2} & {A}_{3} \\ {B}_{1} & {B}_{2} & {B}_{3}\end{matrix}\right]$
=(A_2B_3−B_2A_3)hati−(A_1B_3−B_1A_3)hatj+(A_1B_2−B_1A_2)hatk.
Inserting given vectors we obtain
$\vec{C} =$$\left[\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & - 1 & 2\end{matrix}\right]$
=(1xx2−(-1)xx1)hati−(2xx2−1xx1)hatj+(2xx(-1)−1xx1)hatk.
=3hati−3hatj−3hatk.

Now unit vector in the direction of $\vec{C}$ is $\frac{\vec{C}}{|} \vec{C} |$
$\therefore | \vec{C} | = \sqrt{{3}^{2} + {\left(- 3\right)}^{2} + {\left(- 3\right)}^{2}}$
$= \sqrt{27}$
$= 3 \sqrt{3}$
Therefore desired unit vector is
1/sqrt3(hati−hatj−hatk)