Question

Find a vector which is perpendicular to both vectors $a$ and $b$ , where, \[\vec a = 2i + 3j + 4k\] and $\vec b = i + 2j + 3k$ ?

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.

Complete step by step answer:
Find the cross product of $\vec a$ and $\vec b$ .
$\vec a \times \vec b$ is the determinant of the matrix $\left( {\begin{array}{*{20}{c}}
  i&j&k \\
  2&3&4 \\
  1&2&3
\end{array}} \right)$
$
   \Rightarrow \vec a \times \vec b = \left( {\begin{array}{*{20}{c}}
  i&j&k \\
  2&3&4 \\
  1&2&3
\end{array}} \right) \\
   \Rightarrow \hat i(3 \times 3 - 2 \times 4) - \hat j(2 \times 3 - 1 \times 4) + \hat k(2 \times 2 - 3 \times 1) \\
   \Rightarrow 1\hat i - 2\hat j - 1\hat k \\
 $
Let $\vec c = \vec a \times \vec b$
$\therefore \vec c = 1\hat i - 2\hat j - 1\hat k$
Hence the vector which is perpendicular to \[\vec a = 2i + 3j + 4k\] and $\vec b = i + 2j + 3k$ is $\vec c = 1\hat i - 2\hat j - 1\hat k$.

Note: The 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 vectors mutually perpendicular to each other.
We can even verify whether the obtained vector is perpendicular or not by doing dot product. We know that if two vectors are perpendicular then the dot product is zero. Therefore, we can say that the obtained vector is perpendicular to both $\vec a$ and $\vec b$ .