11.2 Properties of cross product

  Basic Vectors

1. $\overrightarrow A \times \overrightarrow B = - \left( {\overrightarrow B \times \overrightarrow A } \right)$

Magnitude is same. The negative sign indicates that the direction is opposite.

Area of parallelogram is given by, $$\overrightarrow A = \overrightarrow P \times \overrightarrow Q $$

3. $\overrightarrow A \times \left( {\overrightarrow B + \overrightarrow C } \right) = \overrightarrow A \times \overrightarrow B + \overrightarrow A \times \overrightarrow C $

4. $\left| {\widehat i} \right| = \left| {\widehat j} \right| = \left| {\widehat k} \right| = 1$
$\widehat i$, $\widehat j$ & $\widehat k$ are mutually perpendicular vectors.

As, $$\overrightarrow A \times \overrightarrow B = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\sin \theta \ \ \widehat n$$ When we take the cross product of two similar unit vectors, we get, $$\widehat i \times \widehat i = \left( 1 \right)\left( 1 \right)\sin 0^\circ \,\widehat n$$$$\widehat i \times \widehat i = 0$$ Similarly, $$\widehat j \times \widehat j = 0$$$$\widehat k \times \widehat k = 0$$

When we take the cross product of two mutually perpendicular vectors are, $$\widehat i \times \widehat j = \widehat k$$$$\widehat j \times \widehat k = \widehat i$$$$\widehat k \times \widehat i = \widehat j$$ and $$\widehat j \times \widehat i = - \widehat k$$$$\widehat k \times \widehat j = - \widehat i$$$$\widehat i \times \widehat k = - \widehat j$$

5. Consider vectors $\overrightarrow A $ and $\overrightarrow B $ are given as,
$$\overrightarrow A = a\widehat i + b\widehat j + c\widehat k$$$$\overrightarrow B = p\widehat i + q\widehat j + r\widehat k$$$$\overrightarrow A \times \overrightarrow B = \left( {br - qc} \right)\widehat i + \left( {pc - ar} \right)\widehat j + \left( {aq - bp} \right)\widehat k$$
Proof: $$\overrightarrow A \times \overrightarrow B = \left( {a\widehat i + b\widehat j + c\widehat k} \right) \times \left( {p\widehat i + q\widehat j + r\widehat k} \right)$$$$\overrightarrow A \times \overrightarrow B = ap\left( {\widehat i \times \widehat i} \right) + aq\left( {\widehat i \times \widehat j} \right) + ar\left( {\widehat i \times \widehat k} \right) + bp\left( {\widehat j \times \widehat i} \right) + bq\left( {\widehat j \times \widehat j} \right) + br\left( {\widehat j \times \widehat k} \right) + cp\left( {\widehat k \times \widehat i} \right) + cq\left( {\widehat k \times \widehat j} \right) + cr\left( {\widehat k \times \widehat k} \right)$$ As, $$\widehat i \times \widehat i = \widehat j \times \widehat j = \widehat k \times \widehat k = 0$$$$\widehat i \times \widehat j = \widehat k$$$$\widehat j \times \widehat k = \widehat i$$$$\widehat k \times \widehat i = \widehat j$$$$\widehat j \times \widehat i = - \widehat k$$$$\widehat k \times \widehat j = - \widehat i$$$$\widehat i \times \widehat k = - \widehat j$$ So, $$\overrightarrow A \times \overrightarrow B = ap\left( 0 \right) + aq\left( {\widehat k} \right) + ar\left( { - \widehat j} \right) + bp\left( { - \widehat k} \right) + bq\left( 0 \right) + br\left( {\widehat i} \right) + cp\left( {\widehat j} \right) + cq\left( { - \widehat i} \right) + cr\left( 0 \right)$$$$\overrightarrow A \times \overrightarrow B = \left( {br - qc} \right)\widehat i + \left( {pc - ar} \right)\widehat j + \left( {aq - bp} \right)\widehat k$$