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If $|\mathbf{a}|=4,|\mathbf{b}|=5,|\mathbf{a}-\mathbf{b}|=3$ and $\theta$ is the angle between the vectors $\mathbf{a}$ and $\mathbf{b}$, then $\tan ^2 \theta=$
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Verified Answer
The correct answer is:
$\frac{9}{16}$
Given that,
$|\mathbf{a}|=4,|\mathbf{b}|=5,|\mathbf{a}-\mathbf{b}|=3$
Now, $|\mathbf{a}-\mathbf{b}|^2=|\mathbf{a}|^2+|\mathbf{b}|^2-2 \mathbf{a} \cdot \mathbf{b}$
$\begin{array}{ll}
\Rightarrow(3)^2=4^2+5^2-2 \mathbf{a} \cdot \mathbf{b} & \Rightarrow 9=16+25-2 \mathbf{a} \cdot \mathbf{b} \\
\Rightarrow 2 \mathbf{a} \cdot \mathbf{b}=32 & \Rightarrow \mathbf{a} \cdot \mathbf{b}=16
\end{array}$
Now, we know that,
$\cos \theta=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \cdot|\mathbf{b}|}=\frac{16}{4 \cdot 5} \quad \cos \theta=\frac{4}{5}$
So, $\tan \theta=\frac{3}{4}$
$\Rightarrow \quad \tan ^2 \theta=\frac{9}{16}$
$|\mathbf{a}|=4,|\mathbf{b}|=5,|\mathbf{a}-\mathbf{b}|=3$
Now, $|\mathbf{a}-\mathbf{b}|^2=|\mathbf{a}|^2+|\mathbf{b}|^2-2 \mathbf{a} \cdot \mathbf{b}$
$\begin{array}{ll}
\Rightarrow(3)^2=4^2+5^2-2 \mathbf{a} \cdot \mathbf{b} & \Rightarrow 9=16+25-2 \mathbf{a} \cdot \mathbf{b} \\
\Rightarrow 2 \mathbf{a} \cdot \mathbf{b}=32 & \Rightarrow \mathbf{a} \cdot \mathbf{b}=16
\end{array}$
Now, we know that,
$\cos \theta=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \cdot|\mathbf{b}|}=\frac{16}{4 \cdot 5} \quad \cos \theta=\frac{4}{5}$
So, $\tan \theta=\frac{3}{4}$
$\Rightarrow \quad \tan ^2 \theta=\frac{9}{16}$
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