Question

If $|a|=4,|b|=5,|a-b|=3$ and $\theta$ is the angle between the vectors $a$ and $b$, then ${\tan }^2\theta =$

• A. $\frac{4}{3}$
• B. $\frac{3}{4}$
• C. $\frac{16}{9}$
• D. $\frac{9}{16}$

Answer 1

Answer: (D)

Given that,
$|a|=4,|b|=5,|a-b|=3$
Now, $|a-b{|}^2=|a{|}^2+|b{|}^2-2a\cdot b$
$\Rightarrow (3{)}^2=4^2+5^2-2a\cdot b$
$\Rightarrow 9=16+25-2a\cdot b$
$\Rightarrow 2a\cdot b=32$
$\Rightarrow a\cdot b=16$
Now, we know that,
$\cos \theta =\frac{a\cdot b}{|a|\cdot |b|}=\frac{16}{4\cdot 5}$
$\cos \theta =\frac{4}{5}$
So, $\tan \theta =\frac{3}{4}$

$\Rightarrow {\tan }^2\theta =\frac{9}{16}$