Question

In a parallelogram ABCD, if $\overrightarrow{AB}=\overrightarrow{a}\mathrm{and}\overrightarrow{BC}=\overrightarrow{b},\mathrm{then}\overrightarrow{AC}=\_\_\_\_\_\_\_\_\_\_\_\_\_\mathrm{and}\overrightarrow{BD}=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

Given:
ABCD is a parallelogram
$\overrightarrow{AB}=\overrightarrow{a}\mathrm{and}\overrightarrow{BC}=\overrightarrow{b}$

$$\begin{gathered} \mathrm{in}∆ABC, \\ \mathrm{using}\mathrm{triangle}\mathrm{law}\mathrm{of}\mathrm{vector}\mathrm{addition}, \\ \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC} \\ \Rightarrow \overrightarrow{AC}=\overrightarrow{a}+\overrightarrow{b} \\ \mathrm{Now}, \\ \mathrm{in}∆ABD, \\ \overrightarrow{AD}=\overrightarrow{BC}=\overrightarrow{b} \\ \mathrm{using}\mathrm{triangle}\mathrm{law}\mathrm{of}\mathrm{vector}\mathrm{addition}, \\ \overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD} \\ \overrightarrow{BD}=\overrightarrow{AD}-\overrightarrow{AB} \\ \Rightarrow \overrightarrow{BD}=\overrightarrow{b}-\overrightarrow{a} \end{gathered}$$


​Hence, $\overrightarrow{AC}=\underline{\overrightarrow{a}+\overrightarrow{b}}\mathrm{and}\overrightarrow{BD}=\underline{\overrightarrow{b}-\overrightarrow{a}}.$