Question

What is the length of the longer diagonal of the parallelogram constructed on $5\overrightarrow{a}+2\overrightarrow{b}\mathrm{and}\overrightarrow{a}-3\overrightarrow{b}$if it is given that $\left|\overrightarrow{a}\right|=2\sqrt{2},\left|\overrightarrow{b}\right|=3$ and the angle between $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$is π/4?
(a) 15
(b) $\sqrt{113}$
(c) $\sqrt{593}$
(d) $\sqrt{369}$

Answer 1

(c) $\sqrt{593}$


$$\begin{gathered} \mathrm{Let}ABCD\mathrm{be}\mathrm{a}\mathrm{parallelogram}\mathrm{in}\mathrm{which} \\ \mathrm{side}\overrightarrow{AB}=\overrightarrow{DC}=5\overrightarrow{a}+2\overrightarrow{b} \\ \mathrm{and}\overrightarrow{AD}=\overrightarrow{BC}=\overrightarrow{a}-3\overrightarrow{b} \\ \text{a}\mathrm{nd}\mathrm{diagonals}\mathrm{are}AC\mathrm{and}BD. \\ \mathrm{Now},\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC} \\ =\left(5\overrightarrow{a}+2\overrightarrow{b}\right)+\left(\overrightarrow{a}-3\overrightarrow{b}\right) \\ =6\overrightarrow{a}-\overrightarrow{b} \\ \therefore \left|\overrightarrow{AC}\right|=\left|6\overrightarrow{a}-\overrightarrow{b}\right| \\ =\sqrt{{\left|6\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\times \left|6\overrightarrow{a}\right|\times \left|\overrightarrow{b}\right|\cos \theta } \\ =\sqrt{36{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-12\times \left|\overrightarrow{a}\right|\times \left|\overrightarrow{b}\right|\cos \frac{\mathrm{\pi }}{4}} \\ =\sqrt{36{\left|2\sqrt{2}\right|}^2+{\left|3\right|}^2-12\times \left|2\sqrt{2}\right|\times \left|3\right|\times \frac{1}{\sqrt{2}}} \\ =\sqrt{288+9-72} \\ =\sqrt{225}=15\mathrm{units} \\ \overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{BD} \\ =-\overrightarrow{AB}+\overrightarrow{BD} \\ =-\left(5\overrightarrow{a}+2\overrightarrow{b}\right)+\left(\overrightarrow{a}-3\overrightarrow{b}\right) \\ =-4\overrightarrow{a}-5\overrightarrow{b} \\ \therefore \left|\overrightarrow{BD}\right|=\left|-4\overrightarrow{a}-5\overrightarrow{b}\right| \\ =\left|4\overrightarrow{a}+5\overrightarrow{b}\right| \\ =\sqrt{{\left|4\overrightarrow{a}\right|}^2+{\left|5\overrightarrow{b}\right|}^2+2\times \left|4\overrightarrow{a}\right|\times \left|5\overrightarrow{b}\right|\cos \theta } \\ =\sqrt{16{\left|\overrightarrow{a}\right|}^2+25{\left|\overrightarrow{b}\right|}^2+40\times \left|\overrightarrow{a}\right|\times \left|\overrightarrow{b}\right|\cos \frac{\mathrm{\pi }}{4}} \\ =\sqrt{16{\left|2\sqrt{2}\right|}^2+25{\left|3\right|}^2+40\times \left|2\sqrt{2}\right|\times \left|3\right|\times \frac{1}{\sqrt{2}}} \\ =\sqrt{128+225+240} \\ =\sqrt{593}\mathrm{units} \\ \mathrm{Therefore},\text{the}\mathrm{larger}\mathrm{diagonal}=\sqrt{593} \end{gathered}$$