Question

If $\overrightarrow{a}$is a unit vector, then find $\left|\overrightarrow{x}\right|$ in each of the following.
(i) $\left(\overrightarrow{x}-\overrightarrow{a}\right)·\left(\overrightarrow{x}+\overrightarrow{a}\right)=8$

(ii) $\left(\overrightarrow{x}-\overrightarrow{a}\right)·\left(\overrightarrow{x}+\overrightarrow{a}\right)=12$

Answer 1

$$\begin{gathered} \text{Given that}\overrightarrow{a}\text{is a unit vector.} \\ \Rightarrow \left|\overrightarrow{a}\right|=1...\left(1\right) \\ \left(\mathrm{i}\right)\left(\overrightarrow{x}-\overrightarrow{a}\right).\left(\overrightarrow{x}+\overrightarrow{a}\right)=8 \\ \Rightarrow {\left|\overrightarrow{x}\right|}^2-{\left|\overrightarrow{a}\right|}^2=8 \\ \Rightarrow {\left|\overrightarrow{x}\right|}^2-1^2=8\left[\text{From (1)}\right] \\ \Rightarrow {\left|\overrightarrow{x}\right|}^2=9 \\ \Rightarrow \left|\overrightarrow{x}\right|=3 \\ \left(\mathrm{ii}\right)\left(\overrightarrow{x}-\overrightarrow{a}\right).\left(\overrightarrow{x}+\overrightarrow{a}\right)=12 \\ \Rightarrow {\left|\overrightarrow{x}\right|}^2-{\left|\overrightarrow{a}\right|}^2=12 \\ \Rightarrow {\left|\overrightarrow{x}\right|}^2-1^2=12\left[\text{From (1)}\right] \\ \Rightarrow {\left|\overrightarrow{x}\right|}^2=13 \\ \Rightarrow \left|\overrightarrow{x}\right|=\sqrt{13} \end{gathered}$$