Question

How do you compute the dot product for $u=3i+2j$ and $v=-2i-3j$?

Answer 1

Hint: We have two vectors namely \[u,v\] with \[\overset{\wedge }{\mathop{i}}\,\] and \[\overset{\wedge }{\mathop{j}}\,\] as unit vectors. Dot product of the two vectors involves multiplying the \[\overset{\wedge }{\mathop{i}}\,\]component of both the vectors together, that is, \[3.(-2)\] and multiplying the \[\overset{\wedge }{\mathop{j}}\,\] component of both the vectors , that is, \[2.(-3)\] and adding them up. Hence, we get the dot product of both the vectors.

Complete step by step solution:
According to the given question, we have been given two vectors \[u\] and \[v\], on which we have to compute the dot product.
Dot product, is also called the scalar product, as it involves multiplication of similar components to give a scalar result (a number).
Also,
\[\overset{\wedge }{\mathop{i}}\,.\overset{\wedge }{\mathop{i}}\,=1\]
\[\overset{\wedge }{\mathop{j}}\,.\overset{\wedge }{\mathop{j}}\,=1\]

The vectors given to us is,
$u=3i+2j$
$v=-2i-3j$
So the dot product of the two vectors is,
\[u.v=(3i+2j).(-2i-3j)\]-----(1)
Here, we will multiply the \[\overset{\wedge }{\mathop{i}}\,\]component of both the vectors together, similarly, we will multiply the \[\overset{\wedge }{\mathop{j}}\,\] component of both the vectors together, we get,
\[\Rightarrow u.v=3.(-2)+2.(-3)\]-----(2)
We will multiply the terms in the equation (2). The first term is 3 multiplied by (-2) which gives us -6. And the second term has 2 multiplied by (-3) which also gives us -6.
So, we have,
\[\Rightarrow u.v=-6+(-6)\]
Now, we have two (-6) to be added up. We get,
\[\Rightarrow u.v=-6-6\]
\[\Rightarrow u.v=-12\]
Therefore, the dot product of the given vectors is \[-12\].

Note: The dot product involves multiplication of an \[\overset{\wedge }{\mathop{i}}\,\] component with another \[\overset{\wedge }{\mathop{i}}\,\] component and is not multiplied with all the components. We saw in the above solution how a dot product works. We also have an operation called the cross product. It follows a different set of procedures to carry out the operation.
Also,
\[\overset{\wedge }{\mathop{i}}\,\times \overset{\wedge }{\mathop{i}}\,=0\]
\[\overset{\wedge }{\mathop{j}}\,\times \overset{\wedge }{\mathop{j}}\,=0\]