Question

If $(a+b)\cdot (a-b)=8$ and $|a|=8|b|$, then

• A. $|a|=\frac{16}{3}\sqrt{\frac{2}{7}}$
• B. $|b|=\frac{2}{3}\sqrt{\frac{2}{7}}$
• C. Both(a) and (b) are true
• D. Both(a) and (b) are false

Answer 1

Answer: (C)

Given, $(a+b)\cdot (a-b)=8$ and $|a|=8|b|$
$\Rightarrow a\cdot a-a\cdot b+b\cdot a-b\cdot b=8$
$\Rightarrow |a{|}^2-|b{|}^2=8(\because a\cdot a=|a{|}^2$ anda $\cdot b=b\cdot a)$
$\Rightarrow (8|b|{)}^2-|b{|}^2=8$
$\Rightarrow 63|b{|}^2=8($ given, $|a|=8|b|)$
$\Rightarrow |b|=\sqrt{\frac{8}{63}}=\frac{2}{3}\sqrt{\frac{2}{7}}$
Also, $|a|=8|b|=8\left(\frac{2}{3}\sqrt{\frac{2}{7}}\right)=\frac{16}{3}\sqrt{\frac{2}{7}}.$