\displaystyle \textbf{1.}\ \text{If }\overrightarrow{a},\overrightarrow{b}\text{ are two vectors inclined at an angle }\theta,\text{ then }\overrightarrow{a}\times\overrightarrow{b}=|\overrightarrow{a}|\ |\overrightarrow{b}|\ \sin\theta\ \overrightarrow{n},\text{ where }\overrightarrow{n}\text{ is a}
\displaystyle \text{unit vector perpendicular to the plane of }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ such that }\overrightarrow{a},\overrightarrow{b},\overrightarrow{n}\text{ form a right handed system.}

\displaystyle \textbf{2.}\ \overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{0}\ \text{iff }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are parallel.}

\displaystyle \textbf{3.}\ \overrightarrow{a}\times\overrightarrow{b}\text{ gives the vector area of the parallelogram having two adjacent sides as }\overrightarrow{a}\text{ and }\overrightarrow{b}.

\displaystyle \textbf{4.}\ \text{The unit vectors perpendicular to the plane of }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are }\pm\frac{\overrightarrow{a}\times\overrightarrow{b}}{|\overrightarrow{a}\times\overrightarrow{b}|}

\displaystyle \textbf{5.}\ \overrightarrow{a}\times\overrightarrow{b}=-(\overrightarrow{b}\times\overrightarrow{a})

\displaystyle \textbf{6.}\ (i)\ m(\overrightarrow{a}\times\overrightarrow{b})=m\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{a}\times m\overrightarrow{b},\ \text{for any scalar }m
\displaystyle (ii)\ m\overrightarrow{a}\times n\overrightarrow{b}=mn(\overrightarrow{a}\times\overrightarrow{b})=n\overrightarrow{a}\times m\overrightarrow{b}=mn\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{a}\times mn\overrightarrow{b}\ \text{for scalars }m,n

\displaystyle \textbf{7.}\ \overrightarrow{a}\times\overrightarrow{b}=\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\end{vmatrix},\ \text{where }\overrightarrow{a}=a_{1}\widehat{i}+a_{2}\widehat{j}+a_{3}\widehat{k}\text{ and }\overrightarrow{b}=b_{1}\widehat{i}+b_{2}\widehat{j}+b_{3}\widehat{k}

\displaystyle \textbf{8.}\ \text{For any two vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}:\ |\overrightarrow{a}\times\overrightarrow{b}|^{2}+(\overrightarrow{a}\cdot\overrightarrow{b})^{2}=|\overrightarrow{a}|^{2}\ |\overrightarrow{b}|^{2}

\displaystyle \textbf{9.}\ \text{For any vector }\overrightarrow{a}:\ |\overrightarrow{a}\times\widehat{i}|^{2}+|\overrightarrow{a}\times\widehat{j}|^{2}+|\overrightarrow{a}\times\widehat{k}|^{2}=2|\overrightarrow{a}|^{2}

\displaystyle \textbf{10.}\ (i)\ \text{Area of }\triangle ABC=\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|=\frac{1}{2}|\overrightarrow{BC}\times\overrightarrow{BA}|=\frac{1}{2}|\overrightarrow{CB}\times\overrightarrow{CA}|
\displaystyle (ii)\ \text{Area of a plane convex quadrilateral }ABCD\text{ is }\frac{1}{2}|\overrightarrow{AC}\times\overrightarrow{BD}|,\ \text{where }AC\text{ and }BD\text{ are diagonal.}

\displaystyle \textbf{11.}\ \text{If }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ are the position vectors of the vertices }A,B,C\text{ of }\triangle ABC,\text{ then}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\left|\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a}\right|
\displaystyle \text{Length of the perpendicular from }C\text{ on }AB=\frac{\left|\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a}\right|}{|\overrightarrow{b}-\overrightarrow{a}|}
\displaystyle \text{Length of the perpendicular from }A\text{ on }BC=\frac{\left|\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a}\right|}{|\overrightarrow{b}-\overrightarrow{c}|}
\displaystyle \text{Length of the perpendicular from }B\text{ on }AC=\frac{\left|\overrightarrow{a}\times\overrightarrow{b}+\overrightarrow{b}\times\overrightarrow{c}+\overrightarrow{c}\times\overrightarrow{a}\right|}{|\overrightarrow{c}-\overrightarrow{a}|}

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