Question

If $|a\times b{|}^2+(a.b{)}^2=p|b{|}^2$ then $p=$

• A. $1$
• B. $|a{|}^2$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$|a{|}^2$

LHS $=|\overline{a}\times \overline{b}{|}^2+|\overline{a}.\overline{b}{|}^2$
LHS $=||\overline{a}||\overline{b}|cos(\overline{a},\overline{b}){|}^2+||\overline{a}||\overline{b}|sin(\overline{a},\overline{b}){|}^2$
LHS $=|\overline{a}{|}^2|\overline{b}{|}^2\left[co{s}^2\right(\overline{a},\overline{b})+si{n}^2(\overline{a},\overline{b}\left)\right]$
LHS $=|\overline{a}{|}^2|\overline{b}{|}^2$
Thus, $l=|\overline{a}{|}^2$