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If $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are position vectors of the vertices of $\triangle A B C$, then $\frac{(\mathbf{a}-\mathbf{c}) \times(\mathbf{b}-\mathbf{a})}{(\mathbf{b}-\mathbf{a}) \cdot(\mathbf{c}-\mathbf{a})}=$
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2478 Upvotes
Verified Answer
The correct answer is:
$\tan A$
$$
\text { } \frac{(\mathbf{a}-\mathbf{c}) \times(\mathbf{b}-\mathbf{a})}{(\mathbf{b}-\mathbf{a}) \cdot(\mathbf{c}-\mathbf{a})}=\frac{\mathbf{C A} \times \mathbf{A B}}{\mathbf{A B} \cdot \mathbf{A C}}
$$
Hence, option (2) is correct.
\text { } \frac{(\mathbf{a}-\mathbf{c}) \times(\mathbf{b}-\mathbf{a})}{(\mathbf{b}-\mathbf{a}) \cdot(\mathbf{c}-\mathbf{a})}=\frac{\mathbf{C A} \times \mathbf{A B}}{\mathbf{A B} \cdot \mathbf{A C}}
$$
Hence, option (2) is correct.
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