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# Evaluate the product $(3\overrightarrow{a}-5\overrightarrow{b}).(2\overrightarrow{a}+7\overrightarrow{b})$ .

Last updated date: 24th Mar 2023
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Hint: We have to evaluate $(3\overrightarrow{a}-5\overrightarrow{b}).(2\overrightarrow{a}+7\overrightarrow{b})$ , so multiply each other in step by step manner. After that, while simplifying use the properties $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{a}$ and $\overrightarrow{a}.\overrightarrow{a}={{\left| \overrightarrow{a} \right|}^{2}}$ . Try it, you will get the answer.
A vector math is defined as mathematical structure. It has many applications in physics and geometry. We know that the location of the points on the coordinate plane can be represented using the ordered pair such as $(x,y)$ . The usage of vectors are very useful in the simplification process of three-dimensional geometry. Along with the term vector, we have heard the term scalar. A scalar actually represents the “real numbers”. In simpler words, a vector of “ $n$ ” dimensions is an ordered collection of n elements called “components“.
\begin{align} & (3\overrightarrow{a}-5\overrightarrow{b}).(2\overrightarrow{a}+7\overrightarrow{b})=3\overrightarrow{a}.2\overrightarrow{a}+3\overrightarrow{a}.7\overrightarrow{b}-5\overrightarrow{b}.2\overrightarrow{a}-5\overrightarrow{b}.7\overrightarrow{b} \\ & =6(\overrightarrow{a}.\overrightarrow{a})+21(\overrightarrow{a}.\overrightarrow{b})-10(\overrightarrow{b}.\overrightarrow{a})-35(\overrightarrow{b}.\overrightarrow{b}) \\ & =6(\overrightarrow{a}.\overrightarrow{a})+21(\overrightarrow{a}.\overrightarrow{b})-10(\overrightarrow{a}.\overrightarrow{b})-35(\overrightarrow{b}.\overrightarrow{b}) \\ & =6(\overrightarrow{a}.\overrightarrow{a})+11(\overrightarrow{a}.\overrightarrow{b})-35(\overrightarrow{b}.\overrightarrow{b}) \\ & =6{{\left| \overrightarrow{a} \right|}^{2}}+11(\overrightarrow{a}.\overrightarrow{b})-35{{\left| \overrightarrow{b} \right|}^{2}} \\ \end{align} …(Using property $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{a}$ and $\overrightarrow{a}.\overrightarrow{a}={{\left| \overrightarrow{a} \right|}^{2}}$ )
Therefore, $(3\overrightarrow{a}-5\overrightarrow{b}).(2\overrightarrow{a}+7\overrightarrow{b})==6{{\left| \overrightarrow{a} \right|}^{2}}+11(\overrightarrow{a}.\overrightarrow{b})-35{{\left| \overrightarrow{b} \right|}^{2}}$ .