
How do you find the dot product of the vectors $ \overrightarrow{v}=5\widehat{i}-2\widehat{j} $ and $ \overrightarrow{w}=3\widehat{i}+4\widehat{j} $ ?
Answer
471.9k+ views
Hint: We first need to discuss the processes of dot product. We explain the rules and multiplication values for the product. We take the binary operations and find the solution for the multiplication.
Complete step-by-step answer:
We have to find the dot product of the vectors $ \overrightarrow{v}=5\widehat{i}-2\widehat{j} $ and $ \overrightarrow{w}=3\widehat{i}+4\widehat{j} $ .
We keep in mind the different characteristics and conditions for dot products.
The dot product of two vectors is always a scalar value. The dot product of two vectors $ \overrightarrow{m} $ and $ \overrightarrow{n} $ can be expressed as \[\overrightarrow{m}.\overrightarrow{n}=\left| \overrightarrow{m} \right|.\left| \overrightarrow{n} \right|\cos \alpha \] where \[\alpha \] is the angle between those two vectors.
In case of vectors in the form of $ \overrightarrow{m}=a\widehat{i}+b\widehat{j}+c\widehat{k} $ and $ \overrightarrow{n}=d\widehat{i}+e\widehat{j}+f\widehat{k} $ , we just have to take the scalar multiplication of coefficients of same directional units as we have $ \widehat{i}.\widehat{j}=\widehat{j}.\widehat{k}=\widehat{i}.\widehat{k}=0 $ . The rotational multiplication also has the same result where $ \widehat{j}.\widehat{i}=\widehat{k}.\widehat{j}=\widehat{k}.\widehat{i}=0 $ .
The multiplication $ \widehat{i}.\widehat{i}=\widehat{j}.\widehat{j}=\widehat{k}.\widehat{k}=1 $ is the only required part in the scalar multiplication.
We express the vectors $ \overrightarrow{v}=5\widehat{i}-2\widehat{j} $ and $ \overrightarrow{w}=3\widehat{i}+4\widehat{j} $ as $ \overrightarrow{v}=5\widehat{i}-2\widehat{j}+0\widehat{k} $ and $ \overrightarrow{w}=3\widehat{i}+4\widehat{j}+0\widehat{k} $ .
The multiplication gives \[\overrightarrow{v}.\overrightarrow{w}=\left( 5\widehat{i}-2\widehat{j}+0\widehat{k} \right)\left( 3\widehat{i}+4\widehat{j}+0\widehat{k} \right)\].
\[\begin{align}
& \overrightarrow{v}.\overrightarrow{w} \\
& =\left( 5\widehat{i}-2\widehat{j}+0\widehat{k} \right)\left( 3\widehat{i}+4\widehat{j}+0\widehat{k} \right) \\
& =5\times 3+4\times \left( -2 \right) \\
& =15-8 \\
& =7 \\
\end{align}\]
Therefore, the dot product of the vectors $ \overrightarrow{v}=5\widehat{i}-2\widehat{j} $ and $ \overrightarrow{w}=3\widehat{i}+4\widehat{j} $ gives 7.
So, the correct answer is “7”.
Note: The difference of the cross product with the dot product is the difference of being vector and scalar result respectively. The angular part also gives the scalar result also.
Complete step-by-step answer:
We have to find the dot product of the vectors $ \overrightarrow{v}=5\widehat{i}-2\widehat{j} $ and $ \overrightarrow{w}=3\widehat{i}+4\widehat{j} $ .
We keep in mind the different characteristics and conditions for dot products.
The dot product of two vectors is always a scalar value. The dot product of two vectors $ \overrightarrow{m} $ and $ \overrightarrow{n} $ can be expressed as \[\overrightarrow{m}.\overrightarrow{n}=\left| \overrightarrow{m} \right|.\left| \overrightarrow{n} \right|\cos \alpha \] where \[\alpha \] is the angle between those two vectors.
In case of vectors in the form of $ \overrightarrow{m}=a\widehat{i}+b\widehat{j}+c\widehat{k} $ and $ \overrightarrow{n}=d\widehat{i}+e\widehat{j}+f\widehat{k} $ , we just have to take the scalar multiplication of coefficients of same directional units as we have $ \widehat{i}.\widehat{j}=\widehat{j}.\widehat{k}=\widehat{i}.\widehat{k}=0 $ . The rotational multiplication also has the same result where $ \widehat{j}.\widehat{i}=\widehat{k}.\widehat{j}=\widehat{k}.\widehat{i}=0 $ .
The multiplication $ \widehat{i}.\widehat{i}=\widehat{j}.\widehat{j}=\widehat{k}.\widehat{k}=1 $ is the only required part in the scalar multiplication.
We express the vectors $ \overrightarrow{v}=5\widehat{i}-2\widehat{j} $ and $ \overrightarrow{w}=3\widehat{i}+4\widehat{j} $ as $ \overrightarrow{v}=5\widehat{i}-2\widehat{j}+0\widehat{k} $ and $ \overrightarrow{w}=3\widehat{i}+4\widehat{j}+0\widehat{k} $ .
The multiplication gives \[\overrightarrow{v}.\overrightarrow{w}=\left( 5\widehat{i}-2\widehat{j}+0\widehat{k} \right)\left( 3\widehat{i}+4\widehat{j}+0\widehat{k} \right)\].
\[\begin{align}
& \overrightarrow{v}.\overrightarrow{w} \\
& =\left( 5\widehat{i}-2\widehat{j}+0\widehat{k} \right)\left( 3\widehat{i}+4\widehat{j}+0\widehat{k} \right) \\
& =5\times 3+4\times \left( -2 \right) \\
& =15-8 \\
& =7 \\
\end{align}\]
Therefore, the dot product of the vectors $ \overrightarrow{v}=5\widehat{i}-2\widehat{j} $ and $ \overrightarrow{w}=3\widehat{i}+4\widehat{j} $ gives 7.
So, the correct answer is “7”.
Note: The difference of the cross product with the dot product is the difference of being vector and scalar result respectively. The angular part also gives the scalar result also.
Recently Updated Pages
Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Why is the cell called the structural and functional class 12 biology CBSE

Sketch the electric field lines in case of an electric class 12 physics CBSE

State and explain Coulombs law in electrostatics class 12 physics CBSE

a Tabulate the differences in the characteristics of class 12 chemistry CBSE
