Question

If $u=a-b$ and $v=a+b$ and $|a|=|b|=2$, then $|u\times v|$ is equal to

Note: $a$ and $b$ are vectors.

• A. $2\sqrt{16-(a\cdot b{)}^2}$
• B. $\sqrt{16-(a\cdot b{)}^2}$
• C. $2\sqrt{4-(a\cdot b{)}^2}$
• D. $2\sqrt{4+(a\cdot b{)}^2}$

Answer 1

Answer: (A)

$2\sqrt{16-(a\cdot b{)}^2}$

$|u\times v|=|(a-b)\times (a+b)|$
$=2|a\times b|$ $(\because a\times a=b\times b=0)$
and $|a\times b{|}^2+(a\cdot b{)}^2=(absin\theta {)}^2+(abcos\theta {)}^2$
$=a^2{b}^2$
$\Rightarrow |a\times b|=\sqrt{a^2{b}^2-(a\cdot b{)}^2}$
So, $|u\times v|=2|a\times b|$
$=2\sqrt{a^2{b}^2-(a\cdot b{)}^2}$
$=2\sqrt{2^2{2}^2-(a\cdot b{)}^2}$
$=2\sqrt{16-(a\cdot b{)}^2}$
$\therefore |a|=|b|=2$