Question

Let $|\overline{a}+\overline{b}|=|\overline{a}-\overline{b}|$. If $|\overline{a}\times \overline{b}|=\lambda \left|\overline{a}\right|$, then $\lambda =$

• A. $\left|\overline{a}\right|$
• B. $\left|\overline{b}\right|$
• C. $1$
• D. $2$

Answer 1

The correct option is B $\left|\overline{b}\right|$

$|\overline{a}+\overline{b}|=|\overline{a}-\overline{b}|$

$|a{|}^2+|b{|}^2+2\overline{a}\cdot \overline{b}=|a{|}^2+|b{|}^2-2\overline{a}\cdot \overline{b}$

$4\overline{a}\cdot \overline{b}=0$

$\overline{a}\cdot \overline{b}=0$

$\Rightarrow \overline{a}$ & $\overline{b}$ are perpendicular to each other.

$|\overline{a}\times \overline{b}|=|\overline{a}||\overline{b}|\sin \theta$

Since $\overline{a}$ & $\overline{b}$ are perpendicular, $\theta ={90}^o$

$=|\overline{a}|\overline{b}|\sin {90}^o$

$=|a||b|$

$\therefore \lambda =|b|$.