If $\overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are position vectors of the vertices of the triangle $ABC$, then $\frac{\left| \left( \overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{c}} ight) imes \left( \overrightarrow{\mathbf{b}} - \overrightarrow{\mathbf{a}} ight) ight|}{\left( \overrightarrow{\mathbf{c}} - \overrightarrow{\mathbf{a}} ight) \bullet \left( \overrightarrow{\mathbf{b}} - \overrightarrow{\mathbf{a}} ight)}\ is\ equal\ to$

Question

If $\overrightarrow{\mathbf{a}},\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are position vectors of the vertices of the triangle $ABC$, then $\frac{\left| \left( \overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{c}} ight) 	imes \left( \overrightarrow{\mathbf{b}} - \overrightarrow{\mathbf{a}} ight) ight|}{\left( \overrightarrow{\mathbf{c}} - \overrightarrow{\mathbf{a}} ight) \bullet \left( \overrightarrow{\mathbf{b}} - \overrightarrow{\mathbf{a}} ight)}\ is\ equal\ to$

Dr. Abubacker · Accepted Answer

(d)
$\frac{\left| \left( \overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{c}} ight) 	imes \left( \overrightarrow{\mathbf{b}} - \overrightarrow{\mathbf{a}} ight) ight|}{\left( \overrightarrow{\mathbf{c}} - \overrightarrow{\mathbf{a}} ight) \bullet \left( \overrightarrow{\mathbf{b}} - \overrightarrow{\mathbf{a}} ight)} = \frac{\left| \overrightarrow{\mathbf{AC}} 	imes \overrightarrow{\mathbf{BA}} ight|}{\overrightarrow{\mathbf{AC}} \bullet \overrightarrow{\mathbf{BA}}}$
$= \frac{||\overrightarrow{\mathbf{AC}}||\overrightarrow{\mathbf{BA}}|in\ A\widehat{\mathbf{n}}|}{\left| \overrightarrow{\mathbf{AC}} ight|\left| \overrightarrow{\mathbf{BA}} ight|\cos A} = 	an A$

Explanation

Reviewed by Dr. Abubacker

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