Question

Let $\left|\overrightarrow{a}\right|=2\sqrt{2},\left|\overrightarrow{b}\right|=3$ and $(\overrightarrow{a}.\overrightarrow{b})=\frac{\pi }{4}$ if a parallelogram is constructed with adjacent sided $\overrightarrow{a}-3\overrightarrow{b}$ and $\overrightarrow{a}+\overrightarrow{b}$ then the longer diagonal is of length

• A. $10$
• B. $8$
• C. $2\sqrt{26}$
• D. $6$

Answer 1

The correct option is C $2\sqrt{26}$

$\overrightarrow{a}\times \overrightarrow{b}=2\sqrt{2}\times 3\times \frac{1}{\sqrt{2}}=0$
The diagonals are $\overrightarrow{p}+\overrightarrow{q}$ and $\overrightarrow{p}-\overrightarrow{q}$ where $\overrightarrow{p}=2\overrightarrow{a}-3\overrightarrow{b}$ and $\overrightarrow{q}=\overrightarrow{a}+\overrightarrow{b}$
${|\overrightarrow{p}+\overrightarrow{q}|}^2={2\overrightarrow{a}-3\overrightarrow{b}+\overrightarrow{a}+\overrightarrow{b}}^2$
$={|3\overrightarrow{a}-2\overrightarrow{b}|}^2$
$=9{\left|a\right|}^2+4{\left|b\right|}^2-12\overrightarrow{a}\overrightarrow{b}$
$=9{\left(2\sqrt{2}\right)}^2+4\times 3^2-12\times 6$
$=9\times 8+4\times 9-12\times 6=36$ on simplification
${|\overrightarrow{p}-\overrightarrow{q}|}^2={\left|\overrightarrow{a}\right|}^2+16\times {\left|\overrightarrow{b}\right|}^2-8\times \overrightarrow{a}\overrightarrow{b}$
$=8+16\times 9-8\times 6=104$
The length of the larger diagonal is $\sqrt{104}=2\sqrt{26}$