Question

How do you compute the dot product to find the magnitude of $u=\text{ }<2,\text{ }-4>$?

Answer 1

Hint: The vector $u=\text{ }<2,\text{ }-4>$ is given in the matrix form. We can write it in the general form as $u=2i-4j$. Now, the dot product of two vectors is given by the equation $a\cdot b=\left| a \right|\left| b \right|\cos \theta $. So for obtaining the magnitude of the given vector, we have to substitute $a=u~$ and $b=u$, the angle $\theta ={{0}^{\circ }}$ and we will obtain $u\cdot u=\left| u \right|\left| u \right|$. For solving the LHS of the equation, we need to use the distributive law of the dot product.

Complete step by step answer:
The vector given in the question is
$u=\text{ }<2,\text{ }-4>$
The above representation of the given vector is the matrix representation. From the above matrix, we can see that there are the two components of the given vector, which are $2$ and $-4$. From our knowledge of the standard form of the vectors, these are the perpendicular components which are respectively parallel to the $i$ and $j$ directions. So the above vector is written in the standard form as
$\Rightarrow u=2i-4j.........(i)$
Now, we know that the dot product of two vectors $a$ and $b$ is given by
$\Rightarrow a\cdot b=\left| a \right|\left| b \right|\cos \theta $
Where $\theta $ is the angle between the two vectors $a$ and $b$. If we take the two vectors as $a=u~$ and $b=u$ then we will have
$\begin{align}
  & \Rightarrow u\cdot u=\left| u \right|\left| u \right|\cos \theta \\
 & \Rightarrow u\cdot u={{\left| u \right|}^{2}}\cos \theta \\
\end{align}$
Where $\theta $ is the angle between $u$ and $u$. Since a vector is parallel to itself, so the angle $\theta $ in this case must be equal to zero. So we substitute $\theta ={{0}^{\circ }}$ in the above equation to get
\[\Rightarrow u\cdot u={{\left| u \right|}^{2}}\cos {{0}^{\circ }}\]
We know that \[\cos {{0}^{\circ }}=1\]. So we get
\[\begin{align}
  & \Rightarrow u\cdot u={{\left| u \right|}^{2}}\left( 1 \right) \\
 & \Rightarrow u\cdot u={{\left| u \right|}^{2}} \\
 & \Rightarrow {{\left| u \right|}^{2}}=u\cdot u \\
\end{align}\]
Now, we substitute (i) in the above equation to get
\[\Rightarrow {{\left| u \right|}^{2}}=\left( 2i-4j \right)\cdot \left( 2i-4j \right)\]
From the distributive rule of the dot product we can write the above equation as
\[\begin{align}
  & \Rightarrow {{\left| u \right|}^{2}}=2i\cdot \left( 2i-4j \right)-4j\cdot \left( 2i-4j \right) \\
 & \Rightarrow {{\left| u \right|}^{2}}=2i\cdot \left( 2i \right)+2i\cdot \left( -4j \right)+\left( -4j \right)\cdot \left( 2i \right)+\left( -4j \right)\cdot \left( -4j \right) \\
 & \Rightarrow {{\left| u \right|}^{2}}=4i\cdot i-8i\cdot j-8j\cdot i+16j\cdot j \\
\end{align}\]
Now, we know that \[i\cdot i=1\], \[i\cdot j=j\cdot i=0\], and \[j\cdot j=1\]. Substituting these above, we get
\[\begin{align}
  & \Rightarrow {{\left| u \right|}^{2}}=4\left( 1 \right)-8\left( 0 \right)-8\left( 0 \right)+16\left( 1 \right) \\
 & \Rightarrow {{\left| u \right|}^{2}}=4+16 \\
 & \Rightarrow {{\left| u \right|}^{2}}=20 \\
\end{align}\]
Finally, taking square roots of both sides of the above equation, we get the magnitude of the given vector as
$\begin{align}
  & \Rightarrow \left| u \right|=\sqrt{20} \\
 & \Rightarrow \left| u \right|=2\sqrt{5} \\
\end{align}$

Hence, we have computed the dot product for finding the magnitude of the given vector which is equal to $2\sqrt{5}$.

Note: We can also solve this question by taking the dot product of the given vector with some other parallel vector. For obtaining a vector parallel to the given vector, the given vector must be multiplied with a non-zero scalar.