Question

How do you find the cross product and verify that the resulting vectors are perpendicular to the given vectors $ < \text{ }3,2,0\text{ } > \text{ }\times \text{ } < \text{ }1,4,0\text{ } > \text{ }$?

Answer 1

Hint: In this problem we need to calculate the vector product of the two given vectors and check whether the obtained vector is perpendicular to both vectors are not. For this we will first calculate the cross product of the given vectors. We will use the cross-product formula which is $\left( \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right)\times \left( \begin{matrix}
   a \\
   b \\
   c \\
\end{matrix} \right)=\left| \begin{matrix}
   i & j & k \\
   x & y & z \\
   a & b & c \\
\end{matrix} \right|$. So, we will calculate the determinant of the above formed $3\times 3$ matrix and simplify the obtained equation to get the result. Now we need to check whether the resultant vector is a perpendicular vector to given vectors. For this we will calculate the dot product of the vectors and check whether the dot product is zero or not. If the dot product is zero, then the resultant is perpendicular to the given vectors.

Complete step by step solution:
Given vectors, $\left( \begin{matrix}
   3 \\
   2 \\
   0 \\
\end{matrix} \right)$, $\left( \begin{matrix}
   1 \\
   4 \\
   0 \\
\end{matrix} \right)$.
Now the cross product of the above two vectors forms the well-known formula $\left( \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right)\times \left( \begin{matrix}
   a \\
   b \\
   c \\
\end{matrix} \right)=\left| \begin{matrix}
   i & j & k \\
   x & y & z \\
   a & b & c \\
\end{matrix} \right|$, will be
$\Rightarrow \left( \begin{matrix}
   3 \\
   2 \\
   0 \\
\end{matrix} \right)\times \left( \begin{matrix}
   1 \\
   4 \\
   0 \\
\end{matrix} \right)=\left| \begin{matrix}
   i & j & k \\
   3 & 2 & 0 \\
   1 & 4 & 0 \\
\end{matrix} \right|$
Simplifying the determinate of above $3\times 3$ matrix, then we will get
$\begin{align}
  & \Rightarrow \left( \begin{matrix}
   3 \\
   2 \\
   0 \\
\end{matrix} \right)\times \left( \begin{matrix}
   1 \\
   4 \\
   0 \\
\end{matrix} \right)=i\left[ 2\left( 0 \right)-0\left( 4 \right) \right]-j\left[ 3\left( 0 \right)-1\left( 0 \right) \right]+k\left[ 3\left( 4 \right)-1\left( 2 \right) \right] \\
 & \Rightarrow \left( \begin{matrix}
   3 \\
   2 \\
   0 \\
\end{matrix} \right)\times \left( \begin{matrix}
   1 \\
   4 \\
   0 \\
\end{matrix} \right)=0i+0j+10k \\
\end{align}$
Hence the cross product of the given two vectors $\left( \begin{matrix}
   3 \\
   2 \\
   0 \\
\end{matrix} \right)$, $\left( \begin{matrix}
   1 \\
   4 \\
   0 \\
\end{matrix} \right)$ is $\left( \begin{matrix}
   0 \\
   0 \\
   10 \\
\end{matrix} \right)$
To check whether the above resultant vector is perpendicular to the given vectors we are going to calculate the dot product of the resultant vector with the given vectors.
The dot product of the vectors $\left( \begin{matrix}
   3 \\
   2 \\
   0 \\
\end{matrix} \right)$, $\left( \begin{matrix}
   0 \\
   0 \\
   10 \\
\end{matrix} \right)$ will be
$\begin{align}
  & \Rightarrow \left( \begin{matrix}
   3 \\
   2 \\
   0 \\
\end{matrix} \right)\centerdot \left( \begin{matrix}
   0 \\
   0 \\
   10 \\
\end{matrix} \right)=\left( \begin{matrix}
   3\times 0 \\
   2\times 0 \\
   0\times 10 \\
\end{matrix} \right) \\
 & \Rightarrow \left( \begin{matrix}
   3 \\
   2 \\
   0 \\
\end{matrix} \right)\centerdot \left( \begin{matrix}
   0 \\
   0 \\
   10 \\
\end{matrix} \right)=\left( \begin{matrix}
   0 \\
   0 \\
   0 \\
\end{matrix} \right) \\
\end{align}$
Now the dot product of the vectors $\left( \begin{matrix}
   1 \\
   4 \\
   0 \\
\end{matrix} \right)$, $\left( \begin{matrix}
   0 \\
   0 \\
   10 \\
\end{matrix} \right)$.
$\begin{align}
  & \Rightarrow \left( \begin{matrix}
   1 \\
   4 \\
   0 \\
\end{matrix} \right)\centerdot \left( \begin{matrix}
   0 \\
   0 \\
   10 \\
\end{matrix} \right)=\left( \begin{matrix}
   1\times 0 \\
   4\times 0 \\
   0\times 10 \\
\end{matrix} \right) \\
 & \Rightarrow \left( \begin{matrix}
   1 \\
   4 \\
   0 \\
\end{matrix} \right)\centerdot \left( \begin{matrix}
   0 \\
   0 \\
   10 \\
\end{matrix} \right)=\left( \begin{matrix}
   0 \\
   0 \\
   0 \\
\end{matrix} \right) \\
\end{align}$
We can see that the dot product of the resultant vector with both the given vectors is zero. So, the resultant vector is perpendicular to both the given vectors.

Note: In the problem they have clearly mentioned to verify the perpendicularity, so we have calculated the dot product. If any one of the dot products is not zero, then we need to recheck our solution and follow the correct method.