Pls explain the answers of all the questions
Pls answer by this night as i have my exams tomorrow morning by 7 am
Topic : Vectors

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} 2. \\ \mathrm{Given} \mathrm{that} \\ \mathrm{Projection} \mathrm{of} \overrightarrow{\mathrm{b}} \mathrm{on} \overrightarrow{\mathrm{a}}=\mathrm{Projection} \mathrm{of} \overrightarrow{\mathrm{a}} \mathrm{on} \overrightarrow{\mathrm{b}} \\ \frac{\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}}{\left|\overrightarrow{\mathrm{a}}\right|}=\frac{\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}}{\left|\overrightarrow{\mathrm{b}}\right|} \\ \Rightarrow \frac{1}{\left|\overrightarrow{\mathrm{a}}\right|}=\frac{1}{\left|\overrightarrow{\mathrm{b}}\right|} \\ \Rightarrow \left|\overrightarrow{\mathrm{a}}\right|=\left|\overrightarrow{\mathrm{b}}\right| \\ \mathrm{Now} \\ \mathrm{Dot} \mathrm{product} \mathrm{of} \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}} \mathrm{and} \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}} \mathrm{is} \mathrm{given} \mathrm{by} \\ \left(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}\right).\left(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}\right) \\ ={\left|\overrightarrow{\mathrm{a}}\right|}^2-{\left|\overrightarrow{\mathrm{b}}\right|}^2 \\ =0 \left[\mathrm{As} \left|\overrightarrow{\mathrm{a}}\right|=\left|\overrightarrow{\mathrm{b}}\right| \right] \\ \mathrm{As} \mathrm{dot} \mathrm{product} \mathrm{is} 0. \\ \Rightarrow \mathrm{Angle} \mathrm{between} \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}} \mathrm{and} \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}} \mathrm{is} \frac{\mathrm{\pi }}{2} \\ 3. \\ \overrightarrow{\mathrm{a}}\times \left(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}\right)=\left(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}\right)\times \overrightarrow{\mathrm{c}} \\ \Rightarrow \overrightarrow{\mathrm{b}}\left(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}}\right)-\overrightarrow{\mathrm{c}}\left(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}\right)=\overrightarrow{\mathrm{b}}\left(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{c}}\right)-\overrightarrow{\mathrm{a}}\left(\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}\right) \\ \Rightarrow \overrightarrow{\mathrm{c}}\left(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}\right)=\overrightarrow{\mathrm{a}}\left(\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}\right) \\ \Rightarrow \overrightarrow{\mathrm{c}}=\frac{\left(\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}\right)}{\left(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}\right)} \overrightarrow{\mathrm{a}} \\ \mathrm{Dot} \mathrm{product} \mathrm{is} \mathrm{scalar}, \mathrm{hence} \\ \Rightarrow \overrightarrow{\mathrm{c}}=\mathrm{k} \overrightarrow{\mathrm{a}} \\ \mathrm{which} \mathrm{means} \overrightarrow{\mathrm{c}} \mathrm{is} \mathrm{parallel} \mathrm{to} \overrightarrow{\mathrm{a}}. \\ \mathrm{Please} \mathrm{ask} \mathrm{one} \mathrm{question} \mathrm{at} \mathrm{a} \mathrm{time}. \end{gathered}$$

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