Question

Determine a unit vector which is perpendicular to both \[A = 2i + j + k\] and $B = i - j + 2k$ ?

Answer 1

Hint: In cross product (or vector product) of two nonzero vectors $\vec a$ and $\vec b$ , the resultant vector is perpendicular to both vectors $\vec a$ and $\vec b$
So we got the hint, to find a vector perpendicular to two nonzero vectors $\vec a$ and $\vec b$ , we have to find the cross product of those two vectors.
Remember that the resultant vector may or may not be a unit vector.
Unit vectors are those vectors whose magnitude is 1.
Therefore, find the unit vector by dividing the vector by its magnitude.

Complete step by step answer:
Step-1: Find the cross product of $\vec a$ and $\vec b$ .
$\hat a \times \hat b$ is the determinant of the matrix $\left( {\begin{array}{*{20}{c}}
  i&j&k \\
  2&1&1 \\
  1&{ - 1}&2
\end{array}} \right)$
$
   \Rightarrow \hat a \times \hat b = \left( {\begin{array}{*{20}{c}}
  i&j&k \\
  2&1&1 \\
  1&{ - 1}&2
\end{array}} \right) \\
   \Rightarrow \hat i(1 \times 2 - ( - 1) \times 1) - \hat j(2 \times 2 - 1 \times 1) + \hat k(2 \times ( - 1) - 1 \times 1) \\
   \Rightarrow 3\hat i - 3\hat j - 3\hat k \\
 $
Let $\vec c = \hat a \times \hat b$
Step-2: Find the unit vector $\hat c$
$\vec c = 3\hat i - 3\hat j - 3\hat k$
Magnitude of $\vec c$
$
  \left| {\vec c} \right| = \sqrt {{{(3)}^2} + {{( - 3)}^2} + {{( - 3)}^2}} \\
   \Rightarrow \left| {\vec c} \right| = \sqrt {27} \\
   \Rightarrow \left| {\vec c} \right| = 3\sqrt 3 \\
 $
And we know that Unit vector is $\hat c = \dfrac{{\vec c}}{{\left| c \right|}}$
$\therefore \hat c = \dfrac{{3\hat i - 3\hat j - 3\hat k}}{{3\sqrt 3 }}$
So, $\hat c = \dfrac{{\hat i}}{{\sqrt 3 }} - \dfrac{{\hat j}}{{\sqrt 3 }} - \dfrac{{\hat k}}{{\sqrt 3 }}$
Hence the Unit vector perpendicular to both $\vec a$ and $\vec b$ is $\dfrac{{\hat i}}{{\sqrt 3 }} - \dfrac{{\hat j}}{{\sqrt 3 }} - \dfrac{{\hat k}}{{\sqrt 3 }}$

Note: The unit vector $\hat i$ is along the direction of the x-axis, the unit vector $\hat j$ is along the direction of the y-axis, and the unit vector $\hat k$ is along the direction of the z-axis. Thus, $\hat i$ , $\hat j$ and $\hat k$ are unit vectors mutually perpendicular to each other.