Question

Let $|\overrightarrow{a}|=7,|\overrightarrow{b}|=11,|\overrightarrow{a}+\overrightarrow{b}|=10\sqrt{3}$ then what is $|\overrightarrow{a}-\overrightarrow{b}|$ equal to.

• A. $2\sqrt{2}$
• B. $2\sqrt{10}$
• C. 5
• D. 10

Answer 1

Answer: (B)

$2\sqrt{10}$

$|\overrightarrow{a}+\overrightarrow{b}|=10\sqrt{3}$

$\Rightarrow \sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta }=10\sqrt{3}$

$\Rightarrow \sqrt{7^2+{11}^2+2\times 7\times 11\cos \theta }=10\sqrt{3}$

squaring both sides

$\Rightarrow 170+154\cos \theta =300$

$\Rightarrow \cos \theta =\frac{130}{154}$

$|\overrightarrow{a}-\overrightarrow{b}|=$

$=\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta }$

$=\sqrt{7^2+{11}^2-2\times 7\times 11\cos \theta }$

$=\sqrt{170-154\cos \theta }$

$\because \cos \theta =\frac{130}{154}$

$=\sqrt{170-154\times \frac{130}{154}}$

$=\sqrt{40}$

$=2\sqrt{10}$