Question

what is component in dot product

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{The}\mathrm{dot}\mathrm{product}\mathrm{between}\mathrm{two}\mathrm{vectors}\overrightarrow{\mathrm{a}}\mathrm{and}\overrightarrow{\mathrm{b}}\mathrm{is} \\ \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=\left|\mathrm{a}\right|\left|\mathrm{b}\right|\mathrm{cos\theta } \\ \mathrm{where}\mathrm{\theta }\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{vectors}\overrightarrow{\mathrm{a}}\mathrm{and}\overrightarrow{\mathrm{b}}. \\ \mathrm{We}\mathrm{can}\mathrm{expand}\mathrm{the}\mathrm{dot}\mathrm{product}\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{components}, \\ \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=\left({\mathrm{a}}_1\hat{\mathrm{i}}+{\mathrm{a}}_2\hat{\mathrm{j}}+{\mathrm{a}}_3\hat{\mathrm{k}}\right).\left({\mathrm{b}}_1\hat{\mathrm{i}}+{\mathrm{b}}_2\hat{\mathrm{j}}+{\mathrm{b}}_3\hat{\mathrm{k}}\right) \\ \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}={\mathrm{a}}_1{\mathrm{b}}_1\hat{\mathrm{i}}.\hat{\mathrm{i}}+{\mathrm{a}}_2{\mathrm{b}}_2\hat{\mathrm{j}}.\hat{\mathrm{j}}+{\mathrm{a}}_3{\mathrm{b}}_3\hat{\mathrm{k}}.\hat{\mathrm{k}}+\left({\mathrm{a}}_1{\mathrm{b}}_2+{\mathrm{a}}_2{\mathrm{b}}_1\right)\hat{\mathrm{i}}.\hat{\mathrm{j}}+\left({\mathrm{a}}_1{\mathrm{b}}_3+{\mathrm{a}}_3{\mathrm{b}}_1\right)\hat{\mathrm{i}}.\hat{\mathrm{k}} \\ +\left({\mathrm{a}}_2{\mathrm{b}}_3+{\mathrm{a}}_3{\mathrm{b}}_2\right)\hat{\mathrm{j}}.\hat{\mathrm{k}} \\ \mathrm{Since}\mathrm{we}\mathrm{know}\mathrm{the}\mathrm{dot}\mathrm{product}\mathrm{of}\mathrm{unit}\mathrm{vectors},\mathrm{we}\mathrm{can}\mathrm{simplify}\mathrm{the}\mathrm{dot}\mathrm{product}\mathrm{formula}\mathrm{to} \\ \overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}={\mathrm{a}}_1{\mathrm{b}}_1+{\mathrm{a}}_2{\mathrm{b}}_2+{\mathrm{a}}_3{\mathrm{b}}_3 \end{gathered}$$