Let $\vec{a}$ , $\vec{b}$ and $\vec{c}$ be 3 vectors.

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

$(\vec{a} \times \vec{b}) \cdot \vec{c}= \text { Scalar triple product }$

$(\vec{a} \times \vec{b}) \times \vec{c}= \text { Vector triple product }$

i) $\vec{a}, \vec{b}$ and $\vec{a} \times \vec{b}$ from RHS of direction.

ii) $|\vec{a} \times \vec{b}|$ is area of the parallelogram with adjecent sides as $\vec{a}$ and $\vec{b}$

Assume $\vec{a}, \vec{b}, \vec{c}$ are 3 non-coplanar vectors

If $\vec{a}, \vec{b}$ & $\vec{c}$ from RHS of directions.

Then $\vec{c}$ must be making acute angle with $\vec{a} \times \vec{b}$

$(\vec{a} \times \vec{b}) \cdot \vec{c}>0$

If $\vec{a}, \vec{b}, \vec{c}$ from LHS of direction then $\vec{c}$ makes obtuse angle

If $\vec{a}, \vec{b}, \vec{c}$ from LHS of then $\vec{c}$ makes obtuse angle with $\vec{a} \times \vec{b}$. in this case,

$(\vec{a} \times \vec{b}) \cdot \vec{c}<0 .$

$(\vec{a} \times \vec{b}) \cdot \vec{c}=|\vec{a} \times \vec{b}| \underbrace{|\vec{c}| \cos \theta}$

$=$ area of the parallelogram $x$ prop. i $\vec{c}$ along $\vec{a} \times \vec{b}$ :

= Volume of the parallelepiped with edges $O A, O B$ and $O C$

of $\vec{a}, \vec{b}, \vec{c}$ from LHS of direction.

$|(\vec{a} \times \vec{b}) \cdot \vec{c}|$ is equal to the vol.

of the parallelepiped where edges are $O A, O B, O C$

$\therefore(\vec{a} \times \vec{b}) \cdot \vec{c}= \pm V, \quad V$ is the volume.

Notation: $(\vec{a} \times \vec{b}) \cdot \vec{c}=[\vec{a} \vec{b} \vec{c}]$

$( \vec{a} , \vec{b} , \vec{c} )$ from RHS

1. If $\vec{a}, \vec{b}, \vec{c}$ form RHS, $\vec{b}, \vec{c}, \vec{a}$ also form RHS.

$\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=\left[\begin{array}{lll}\vec{b} & \vec{c} & \vec{a}\end{array}\right]=\left[\begin{array}{lll}\vec{c} & \vec{a} & \vec{b}\end{array}\right]$

2. $[\hat{i} \hat{j} \hat{k}]=1=[j k i]=[k i j]$ and

$[\hat{i} \hat{j} \hat{k}] = -1 = [j i k]=[k j i]$

How to Calculate $(\vec{a} \times \vec{b}) \cdot \vec{c}$.

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

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

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

$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3\end{array}\right|$

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

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

$(\vec{a} \times \vec{b}) \cdot \vec{c}$

$= c_1\left(a_2 b_3-b_2 a_3\right)-c_2\left(a_1 b_3-a_3 b_1\right) +c_3\left(a_1 b_2-b_1 a_2\right)$ =

$\left|\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right|$

Given $\vec{a} \vec{b}, \vec{c}$

1. Suppose one of them is zero vector

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

2. Suppose two of the $\vec{a}, \vec{b}, \vec{c}$ are parallel or coincident

(2). Suppose two of the 0,0 , parallel or coincident Then also $(\vec{a} \times \vec{b}) \cdot \vec{c}=0$

(3) Suppose then assumption in (1) & (2) are not true.

Then $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=0 \Rightarrow \vec{a}, \vec{b}, \vec{c}$ are coplanar conversly,

if $\vec{a}, \vec{b}, \vec{c}$ are coplanar, (ie) $\vec{c}=\alpha \vec{a}+\beta \vec{b}$

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

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

$=0+0=0$

Ex. $\vec{a}=i+2 j-k,$ $\vec{b}=2 i+3 j+4 k,$

$\vec{c}=3 i+4 j+9 k$

Ex:

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

$\vec{b} =2 i+3 j+4 k$

$\vec{c} =3 i+4 j+9 k$ ] are coplanar

$(\vec{a} \times \vec{b}) \cdot \vec{c}$

$=\left|\begin{array}{ccc}1 & 2 & -1 \\ 2 & 3 & 4 \\ 3 & 4 & 9\end{array}\right|=0$

$\vec{a}=2 i-3 j+4 k$

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

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

Find $\lambda$ so that these are coplanar. $\lambda = \frac{8}{5}$

$(\vec{a} \times \vec{b}) \times \vec{c} ?$

$\text { Let } \vec{d}=(\vec{a} \times \vec{b}) \times \vec{c}$

$\vec{d} \perp \times(\vec{a} \times \vec{b}) \ \vec{d} \perp r \vec{c} .$

But $\vec{a} \times \vec{b}$ is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$. That means, $\vec{d}$ lies on the plane containing $\vec{a}$ and $\vec{b}$.

Then $\vec{d}=\alpha \vec{a}+\beta \vec{b}$.

Then

$\vec{d}=\alpha \vec{a}+\beta \vec{b}$

$\vec{d} \cdot \vec{c}=\alpha \vec{a} \cdot \vec{c}+\beta \vec{b} \cdot \vec{c}$

$0=\alpha \vec{a} \cdot \vec{c}+\beta \vec{b} \cdot \vec{c}$

$\frac{\alpha}{\vec{b} \cdot \vec{c}}=-\frac{\beta}{\vec{a} \cdot \vec{c}}(=\lambda)$

$\Rightarrow \alpha=\lambda \vec{b} \cdot \vec{c}$ and $\beta=-\lambda(\vec{a} \cdot \vec{c})$

$\therefore(\vec{a} \times \vec{b}) \times \vec{c}=\lambda[(\vec{b} \cdot \vec{c}) \vec{a}-(\vec{a} \cdot \vec{c}) \vec{b}]$

Claim: $-\lambda=-1$.

Given $\vec{a}, \vec{b}, \vec{c}$

Choose the unit vectors $i, j$ \& $k$ mutually $\perp$ \& form RHS as follows:

choose $i$ along $\vec{a}$

(ii) $\vec{a}=a_1 i$

choose $j$ in the plane of $\vec{a}$ and $\vec{b}$ so that it is $\perp_r$ to $i$. So that it is Ir

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

Choose $k$ Iv to the plane of $\vec{a} \& \vec{b}$

so that $i, j, k$ form RHS.

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

$(\vec{a} \times \vec{b}) \times \vec{c}=a_1 b_2 c_1 j-a_1 b_2 c_2 i$

Then

$(\vec{a} \times \vec{b}) \times \vec{c}=a_1 b_2 c_1 j-a_1 b_2 c_2 i$

$(\vec{b} \cdot \vec{c}) \vec{a}-(\vec{a} \cdot \vec{c}) \vec{b}=a_1 b_2 c_2 i-a_1 b_2 c_1 j$

$\Rightarrow \lambda=-1$

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

Note :-

$(\vec{a} \times \vec{b}) \times \vec{c} \neq \vec{a} \times(\vec{b} \times \vec{c})$

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

Ex.

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

$|\vec{b}| \cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$ (Projection)

The projection vector is $\frac{(\vec{a} \cdot \vec{b}) \vec{a}}{|\vec{a}|^2}$.

The perpendicular component of $b$ is given by

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

Aim:- To find a formula $\rightarrow{* *}$

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

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

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

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

$= \frac{\vec{a} \times(\vec{b} \times \vec{a})}{|\vec{a}|^2}$