Question

If $|\overrightarrow{a}+\overrightarrow{b}|=60,|\overrightarrow{a}-\overrightarrow{b}|=40,$ and $|\overrightarrow{b}|=46$. Find $|\overrightarrow{a}|$

• A. 11
• B. 22
• C. 26
• D. 32

Answer 1

The correct option is B 22

Here, $|\overrightarrow{a}+\overrightarrow{b}|=60$
Squaring both the sides,
$|\overrightarrow{a}+\overrightarrow{b}{|}^2=(60{)}^2$
$|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+2\overrightarrow{a}.\overrightarrow{b}=3600$ ...(i)
Now, $|\overrightarrow{a}-\overrightarrow{b}|=40$
Squaring both the sides,
$|\overrightarrow{a}-\overrightarrow{b}{|}^2=(40{)}^2$
$|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2-2\overrightarrow{a}.\overrightarrow{b}=1600$ ...(ii)

Adding (i) and (ii)
$2|\overrightarrow{a}{|}^2+2|\overrightarrow{b}{|}^2+2\overrightarrow{a}.\overrightarrow{b}-2\overrightarrow{a}.\overrightarrow{b}=3600+1600$
$2|\overrightarrow{a}{|}^2+2\times {46}^2=5200$
$2|\overrightarrow{a}|=5200-4232$
$2|\overrightarrow{a}{|}^2=968$
$|\overrightarrow{a}{|}^2=\frac{968}{2}$
$|\overrightarrow{a}{|}^2=484$
$|\overrightarrow{a}|=\sqrt{484}$
$|\overrightarrow{a}|=22$