Lecture 4 MATH 127 (Maj Sultana)

MATH 127 

Vector Analysis

Triple Product

1.  $\vec{A} .(\vec{B} \times \vec{C} )$  

2.  $\vec{A} \times (\vec{B} \times \vec{C} )$  

3.  $(\vec{A} .\vec{B} ). \vec{C}$  

    • Projection: 

i) Scaler Projection:  $\displaystyle \vec{b} \frac{\vec{a} }{|\vec{a}| }$  

ii) Vector Projection:  $\displaystyle \frac{\vec{a} .\vec{b} }{|\vec{a}|^2} \vec{a}$  

    The scaler projection of  $\vec{b}$ and  $\vec{a}$ is the length of the AB (shown in the figure). The vector projection of  $\vec{b}$ onto  $\vec{a}$ is the vector with this length at the point A point in the same direction as  $\vec{a}$ .

    This quantity is also called the component of  $\vec{b}$ in the  $\vec{a}$ direction. And, the vector projection is merely the unit vector  $\displaystyle \frac{\vec{a} }{|\vec{a} }$ times the scalar projection of  $\vec{b}$ onto  $\vec{a}$ .

$\therefore$ The scalar projection of  $\vec{b}$ onto  $\vec{a}$ is proj  $\displaystyle \frac{\vec{b} }{\vec{a}} = \vec{b} .\frac{\vec{a}}{\vec{a}}$  

and the vector projection of    $\vec{b}$ onto  $\vec{a}$ is proj  $\displaystyle \frac{\vec{b} }{\vec{a}} = \frac{\vec{a} .\vec{b} }{|\vec{a} |^2} \vec{a}$ .

    ♦ Direction Cosines: 

  $$\cos \alpha = \frac{x}{|\vec{r}|} = \frac{x}{\sqrt{ x^2 + y^2 + z^2}}$$  

  $$\cos \beta = \frac{y}{|\vec{r}|} = \frac{y}{\sqrt{x^2 + y^2 + z^2}}$$  

  $$\cos \gamma = \frac{z}{|\vec{r}|} = \frac{z }{\sqrt{x^2 + y^2 + z^2} }$$  

Theorem:  $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$  

    Example: Find direction cosines of a vector represented by  $\vec{PQ }$ where  $P(2,- 3, 5) \textrm{ and } Q(1,0,- 1 )$ are two point.

Solution:  $$\begin{align*} \vec{PQ} & = (1 - 2) \hat{i} + (0 + 3) \hat{j} + (- 1 - 5) \hat{k} \\ & = - \hat{i} + 3 \hat{j} - 5 \hat{k} \\ \textrm{Now,} & \\ \alpha & = \cos^{-1} \left( \frac{- 1}{\sqrt[]{1^2 + 3^2 + 6^2}}\right) \\ & = \cos^{-1} \left(- \frac{1}{\sqrt{46}}\right) ~= 98.47^\circ \\ \beta & = \cos^{-1} \left(\frac{3}{\sqrt{1^2 + 3^2 + 6^2}}\right) \\ & = \cos^{-1} \frac{3}{\sqrt{46}} ~= 63.74^\circ \\ \theta & = \cos^{-1} \left(\frac{- 6}{\sqrt[]{1^2 + 3^2 + 6^2}}\right) \\ & = \cos^{-1} \frac{6}{\sqrt{46}} ~= 152.2^\circ \end{align*}$$