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Q.
If $| a |=4,| b |=5,| a - b |=3$ and $\theta$ is the angle between the vectors $a$ and $b$, then $\tan ^{2} \theta=$
TS EAMCET 2018
Solution:
Given that,
$| a | =4, | b |=5,| a - b |=3 $
Now, $| a - b |^{2}=| a |^{2}+| b |^{2}-2 a \cdot b$
$\Rightarrow \, (3)^{2}=4^{2}+5^{2}-2 a \cdot b$
$ \Rightarrow \,9=16+25-2 a \cdot b$
$\Rightarrow \,2 a \cdot b =32$
$\Rightarrow \, a \cdot b =16$
Now, we know that,
$\cos\, \theta=\frac{ a \cdot b }{| a | \cdot| b |}=\frac{16}{4 \cdot 5} $
$ \cos \,\theta=\frac{4}{5}$
So, $\tan\, \theta=\frac{3}{4}$
$\Rightarrow \, \tan ^{2} \theta=\frac{9}{16}$
