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Question: How do you find the dot product of the vectors \( \overrightarrow{v}=5\widehat{i}-2\widehat{j} \) an...

How do you find the dot product of the vectors v→=5i^−2j^\overrightarrow{v}=5\widehat{i}-2\widehat{j} and w→=3i^+4j^\overrightarrow{w}=3\widehat{i}+4\widehat{j} ?

Explanation

Solution

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 v→=5i^−2j^\overrightarrow{v}=5\widehat{i}-2\widehat{j} and w→=3i^+4j^\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 m→\overrightarrow{m} and n→\overrightarrow{n} can be expressed as m→.n→=∣m→∣.∣n→∣cos⁡α\overrightarrow{m}.\overrightarrow{n}=\left| \overrightarrow{m} \right|.\left| \overrightarrow{n} \right|\cos \alpha where α\alpha is the angle between those two vectors.
In case of vectors in the form of m→=ai^+bj^+ck^\overrightarrow{m}=a\widehat{i}+b\widehat{j}+c\widehat{k} and n→=di^+ej^+fk^\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 i^.j^=j^.k^=i^.k^=0\widehat{i}.\widehat{j}=\widehat{j}.\widehat{k}=\widehat{i}.\widehat{k}=0 . The rotational multiplication also has the same result where j^.i^=k^.j^=k^.i^=0\widehat{j}.\widehat{i}=\widehat{k}.\widehat{j}=\widehat{k}.\widehat{i}=0 .
The multiplication i^.i^=j^.j^=k^.k^=1\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 v→=5i^−2j^\overrightarrow{v}=5\widehat{i}-2\widehat{j} and w→=3i^+4j^\overrightarrow{w}=3\widehat{i}+4\widehat{j} as v→=5i^−2j^+0k^\overrightarrow{v}=5\widehat{i}-2\widehat{j}+0\widehat{k} and w→=3i^+4j^+0k^\overrightarrow{w}=3\widehat{i}+4\widehat{j}+0\widehat{k} .
The multiplication gives v→.w→=(5i^−2j^+0k^)(3i^+4j^+0k^)\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).

& \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{aligned}$$ 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.