Vector Product (cross product)

Right handed system of directions OX, OY, OZ

$D X, O Y, O Z^{\prime}$ is left handed system.

$O A, O B, O C$ RHS of directions.

$O A, O B, O C^{\prime}$ LHS of directions.

Let $\vec{a}$ & $\vec{b}$ be two vectors. Let$\theta$ be the angle between them and

$0 \leq \theta \leq \pi$.

$\vec{a} \times \vec{b}:=|\vec{a}||\vec{b}| \hat{n} \sin \theta$

where $\hat{n}$ is a unit vector such that at 0 ,

$\vec{a}, \vec{b}, \hat{n}$ (in this specific order)

$\vec{a} \cdot \vec{b}=|a||b| \cos \theta$

form a RHS of directed

$\quad \vec{a} \times \vec{b}$

i) $\vec{a}, \vec{b}, \hat{n}$ from a RHS of directions

ii) $|\vec{a} \times \vec{b}|=|\vec{a}| \cdot|\vec{b}| \sin \theta$

iii) $\vec{a} \times \vec{b}$ is perpendicular to each of $\vec{a} \times \vec{b}$.

consider $\vec{a}$ & $\vec{b}$

$\text { Area of parallelogram }$

$=\text { base } \times \text { height }$

$=(a) \times|b| \sin \theta$ .

$|\vec{a} \times \vec{b}|$

$=\text { Area } of \text { the }$ parallelogram whose adjustment sides are $\vec{a}$ & $\vec{b}$.

Also,$\frac{1}{2}|\vec{a} \times \vec{b}|$ gives

the area of $\triangle A B C$ whose adjustment sides are $\vec{a}$ & $\vec{b}$.

i) If one of $\vec{a}$ & $\vec{b}$ is $\overrightarrow{0}, \vec{a} \times \vec{b}=\overrightarrow{0}$

ii) If $\vec{a}$ & $\vec{b}$ are parallel,

$\vec{a} \times \vec{b}=\overrightarrow{0} \quad(\because \theta=0, {or}\quad \pi)$

iii) $\alpha \vec{a} \times \beta \vec{b}=\alpha \beta(\vec{a} \times \vec{b})=\vec{a} \times(\alpha \beta \vec{b})$

iv) $\vec{a} \times \vec{b}=-\vec{b} \times \vec{a}$

v) $\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$.

vi)

$|\vec{a} \times \vec{b}|^2 =|\vec{a}|^2|\vec{b}|^2 \cdot \sin ^2 \theta$

$=|\vec{a}|^2|\vec{b}|^2\left(1-\cos ^2 \theta\right)$

$=|\vec{a}|^2|\vec{b}|^2-|\vec{a}|^2|\vec{b}|^2 \cos ^2 \theta$

$[ \vec{a} \cdot \vec{b}=|\vec{a}|| \vec{b} \mid \cdot \cos \theta ]$

$=|\vec{a}|^2|\vec{b}|^2-(\vec{a} \cdot \vec{b})^2$

${\text {Vii) }} (\vec{a}+\vec{b}) \times(\vec{c}+\vec{d})$

$=\vec{a} \times \vec{c}+\vec{a} \times \vec{d}+\vec{b} \times \vec{c}+\vec{b} \times \vec{d}$

To compute$\quad \vec{a} \times \vec{b}.$ Let,

$\text { Let } \vec{a}=a_1 i+a_2 j+a_3 k$ $\quad \quad \vec{b}=b_1 i+b_2 j+b_3 k$

$\vec{a} \times \vec{b}=\left(a_1 i+a_2 j+a_3 k\right) \times$

$\left(b_1 i+b_2 j+b_3 k\right)$

$i \times i=0=j \times j=k \times k,$

$i \times j=k,$

$j \times k=i,$

$k \times i=j,$

$i \times k=-j,$

$j \times i=-k,$

$k \times j=-i,$

$= a_1 b_1 i \times i+a_1 b_2 i \times j+a_1 b_3 i \times k$

$+ a_2 b_1 j \times i+a_2 b_2 j \times j+a_2 b_3 j \times k$

$+ a_3 b_1 k \times i+a_3 b_2 k \times j+a_3 b_3 k \times k$

$= a_1 b_2 k-a_1 b_3 j \ -a_2 b_1 k+a_2 b_3 c \ +a_3 b_1 j-a_3 b_2 i$

$= \left(a_2 b_3-a_3 b_2\right) i+\left(a_3 b_1-a_1 b_3\right) j+$

$\left(a_1 b_2-a_2 b_1\right) k$

$\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$

Proof.

$\vec{a} =a_1 i+a_2 j+a_3 k$

$\vec{b} =b_1 i+b_2 j+b_3 k$

$\vec{c} =c_1 i+c_2 j+c_3 k$

$\vec{a} \times(\vec{b}+\vec{c})=\left|\begin{array}{ccc}i & j & k \\ a_1 & a_2 & a_3 \\ b_1+c_1 & b_2+c_2 & b_3+c_3\end{array}\right|$

$\begin{aligned} & =\left|\begin{array}{ccc}i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3\end{array}\right|+\left|\begin{array}{ccc}i & j & k \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3\end{array}\right| \\ & =\vec{a} \times \vec{b}+\vec{a} \times \vec{c} .\end{aligned}$

From the definition. of $\vec{a} \times \vec{b}$

$\hat{n}=\frac{\vec{a} \times \vec{b}}{|\vec{a}||\vec{b}| \sin \theta}=\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$

$\text { and } \sin \theta=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}$ .

$\begin{aligned}& \vec{a}=3 i-6 j+2 k \& \vec{b}=i-2 j-2 k\end{aligned}$

Find a unit vector perpendicular to both $\vec{a}$ & $\vec{b}$ and find the angle between $\vec{a}$ & $\vec{b}$.

$\hat{n}=\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$

$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}i & j & k \\ 3 & -6 & 2 \\ 1 & -2 & -2\end{array}\right|=16 i+8 j$

$|\vec{a} \times \vec{b}|=\sqrt{256+64}=\sqrt{320}=8 \sqrt{5}$

$\therefore$ The required unit vector $=\frac{8(2 i+j)}{8 \sqrt{5}}=\frac{1}{\sqrt{5}}(2 i+j)$

Here, $|\vec{a}|=7$ & $|\vec{b}|=3$.

$\sin \theta =\frac{|\vec{a} \times \vec{b}|}{7 \times 3}=\frac{8 \sqrt{5}}{21}$

$\theta =\sin ^{-1}\left(\frac{8 \sqrt{5}}{21}\right)$

Take

$\vec{a}=2 i-j+2 k$

$\vec{b}=10 i-2 j+7 k$

Let $|\vec{a}|=10,|\vec{b}|=2, \vec{a} \cdot \vec{b}=12$

Find $|\vec{a} \times \vec{b}|$

$|\vec{a} \times \vec{b}|^2 =|\vec{a}|^2\left|b^2\right|-(\vec{a} \cdot \vec{b})^2$

$=400-144=256$

$\therefore|\vec{a} \times \vec{b}| =16:$

For any 3 vectors $\vec{a}, \vec{b}$ and $\vec{c}$

Prove that $\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times(\vec{c}+\vec{a})+\vec{c} \times(\vec{a}+\vec{b})=\overrightarrow{0}$

Ex. Given $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$

Prove that $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}$.

$\vec{a} \times(\vec{a}+\vec{b}+\vec{c})=\overrightarrow{0}$

$\vec{a} \times \vec{b}+\vec{a} \times \vec{c}=0$

$\therefore \vec{a} \times \vec{b}=\vec{c} \times \vec{a}$

Similarly, $\vec{b} \times(\vec{a}+\vec{b}+\vec{c})=0$

gives $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}$