Question

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

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

Answer 1

Answer: (A)

$|u\times v|=|(a-b)\times (a+b)|$
$=2|a\times b|$
$(\because a\times a=b\times b=0)$
and $|a\times b{|}^2+(a\cdot b{)}^2$
$=(ab\sin \theta {)}^2+(ab\cos \theta {)}^2$
$=a^2{b}^2$
$\Rightarrow |a\times b|=\sqrt{a^2{b}^2-(a\cdot b{)}^2}$
So, $|u\times v|=2|a\times b|$
$=2\sqrt{a^2{b}^2-(a\cdot b{)}^2}$
$=2\sqrt{2^2{2}^2-(a\cdot b{)}^2}$
$=2\sqrt{16-(a\cdot b{)}^2}$
$\left(\because |a|=|b|=2\right)$