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Question: It is given that we have two vectors such as \[\overrightarrow{u}=2\overrightarrow{i}+3\overrightarr...

It is given that we have two vectors such as u=2i+3j\overrightarrow{u}=2\overrightarrow{i}+3\overrightarrow{j} and v=3i+2j\overrightarrow{v}=3\overrightarrow{i}+2\overrightarrow{j}. How to calculate uv\overrightarrow{u}\cdot \overrightarrow{v}?

Explanation

Solution

This question is from the topic of vector and 3-dimensional geometry. In this question, we will find the dot product of the vectors u and v. For finding the dot product, we will first know the formula of dot product, and then apply that formula in solving this question. After solving the further question, we will get our answer.

Complete step-by-step solution:
Let us solve this question.
In this question, we have asked to find out the value of uv\overrightarrow{u}\cdot \overrightarrow{v} using the given data in the question.
The given data are:
u=2i+3j\overrightarrow{u}=2\overrightarrow{i}+3\overrightarrow{j}
v=3i+2j\overrightarrow{v}=3\overrightarrow{i}+2\overrightarrow{j}
The formulas for the dot product are:
ij=ji=0\overrightarrow{i}\cdot \overrightarrow{j}=\overrightarrow{j}\cdot \overrightarrow{i}=0
ii=jj=1\overrightarrow{i}\cdot \overrightarrow{i}=\overrightarrow{j}\cdot \overrightarrow{j}=1
Whenever we have to do the dot product of two vectors, we will simply multiply the vectors.
So, we can write
uv=(2i+3j)(3i+2j)\overrightarrow{u}\cdot \overrightarrow{v}=\left( 2\overrightarrow{i}+3\overrightarrow{j} \right)\cdot \left( 3\overrightarrow{i}+2\overrightarrow{j} \right)
We will use foil method here. The formula for foil method is: (a+b)(c+d)=ac+ad+bc+bd\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd
Using this formula, we can the above equation as
uv=(2i+3j)(3i+2j)=(2i)(3i)+(2i)(2j)+(3j)(3i)+(3j)(2j)\Rightarrow \overrightarrow{u}\cdot \overrightarrow{v}=\left( 2\overrightarrow{i}+3\overrightarrow{j} \right)\cdot \left( 3\overrightarrow{i}+2\overrightarrow{j} \right)=\left( 2\overrightarrow{i} \right)\cdot \left( 3\overrightarrow{i} \right)+\left( 2\overrightarrow{i} \right)\cdot \left( 2\overrightarrow{j} \right)+\left( 3\overrightarrow{j} \right)\cdot \left( 3\overrightarrow{i} \right)+\left( 3\overrightarrow{j} \right)\cdot \left( 2\overrightarrow{j} \right)
We can write the above equation as
uv=(2i+3j)(3i+2j)=6(i)(i)+4(i)(j)+9(j)(i)+6(j)(j)\Rightarrow \overrightarrow{u}\cdot \overrightarrow{v}=\left( 2\overrightarrow{i}+3\overrightarrow{j} \right)\cdot \left( 3\overrightarrow{i}+2\overrightarrow{j} \right)=6\left( \overrightarrow{i} \right)\cdot \left( \overrightarrow{i} \right)+4\left( \overrightarrow{i} \right)\cdot \left( \overrightarrow{j} \right)+9\left( \overrightarrow{j} \right)\cdot \left( \overrightarrow{i} \right)+6\left( \overrightarrow{j} \right)\cdot \left( \overrightarrow{j} \right)
The above equation can also be written as
uv=(2i+3j)(3i+2j)=6(ii)+4(ij)+9(ji)+6(jj)\Rightarrow \overrightarrow{u}\cdot \overrightarrow{v}=\left( 2\overrightarrow{i}+3\overrightarrow{j} \right)\cdot \left( 3\overrightarrow{i}+2\overrightarrow{j} \right)=6\left( \overrightarrow{i}\cdot \overrightarrow{i} \right)+4\left( \overrightarrow{i}\cdot \overrightarrow{j} \right)+9\left( \overrightarrow{j}\cdot \overrightarrow{i} \right)+6\left( \overrightarrow{j}\cdot \overrightarrow{j} \right)
Now, using the formulas: ij=ji=0\overrightarrow{i}\cdot \overrightarrow{j}=\overrightarrow{j}\cdot \overrightarrow{i}=0 and ii=jj=1\overrightarrow{i}\cdot \overrightarrow{i}=\overrightarrow{j}\cdot \overrightarrow{j}=1, we can write the above equation as
uv=(2i+3j)(3i+2j)=6(1)+4(0)+9(0)+6(1)\Rightarrow \overrightarrow{u}\cdot \overrightarrow{v}=\left( 2\overrightarrow{i}+3\overrightarrow{j} \right)\cdot \left( 3\overrightarrow{i}+2\overrightarrow{j} \right)=6\left( 1 \right)+4\left( 0 \right)+9\left( 0 \right)+6\left( 1 \right)
The above equation can also be written as
uv=(2i+3j)(3i+2j)=6+0+0+6\Rightarrow \overrightarrow{u}\cdot \overrightarrow{v}=\left( 2\overrightarrow{i}+3\overrightarrow{j} \right)\cdot \left( 3\overrightarrow{i}+2\overrightarrow{j} \right)=6+0+0+6
The above equation can also be written as
uv=(2i+3j)(3i+2j)=12\Rightarrow \overrightarrow{u}\cdot \overrightarrow{v}=\left( 2\overrightarrow{i}+3\overrightarrow{j} \right)\cdot \left( 3\overrightarrow{i}+2\overrightarrow{j} \right)=12
Hence, we calculated the value of uv\overrightarrow{u}\cdot \overrightarrow{v} using the given data in the question. The calculated value of uv\overrightarrow{u}\cdot \overrightarrow{v} is 12.

Note: We should have a better knowledge in the topic of vector and 3-dimensional geometry to solve this type of question easily. Remember the following formulas whenever we have to do the dot product:
uv=uvcosθ\overrightarrow{u}\cdot \overrightarrow{v}=\left| u \right|\left| v \right|\cos \theta , where θ\theta is the angle between the vectors u and v
ij=ji=0\overrightarrow{i}\cdot \overrightarrow{j}=\overrightarrow{j}\cdot \overrightarrow{i}=0
ii=jj=1\overrightarrow{i}\cdot \overrightarrow{i}=\overrightarrow{j}\cdot \overrightarrow{j}=1
If A=a1i+a2j\overrightarrow{A}={{a}_{1}}\overrightarrow{i}+{{a}_{2}}\overrightarrow{j} and B=b1i+b2j\overrightarrow{B}={{b}_{1}}\overrightarrow{i}+{{b}_{2}}\overrightarrow{j}, then their dot product will be:
AB=a1b1+a2b2\overrightarrow{A}\cdot \overrightarrow{B}={{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}
Also remember the foil method formula which is in the following:
(a+b)(c+d)=ac+ad+bc+bd\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd