Lecture 3 [11 Feb, 2020] MATH 127 (Maj Sultana)

 MATH 127

Vector Analysis

    • Find the angle between two vectors where  $\vec{A} = 2 \hat{i} + 2 \hat{j} \textrm{ and } \vec{B} = 6 \hat{i} - 3 \hat{j}$  

  $$\begin{align*} \theta & = \cos^{-1} \frac{\vec{A}. \vec{B} }{|\vec{A} |. |\vec{B} |} \\ & = \cos^{-1} \frac{12 - 6}{2 \sqrt{2}. \sqrt{45}}\\ & = 71.56^\circ \end{align*}$$  

    • Important properties: 

1. The area of the parallelogram is given,  A =  $|\vec{A} \times \vec{B}|$  

2. The volume of the parallelepiped is given, V =    $|\vec{A} .(\vec{B} \times \vec{C} |)$  

3. Are of Triangles =  $\frac{1}{2} |\vec{A} \times \vec{B} |$  

4. If the three vectors a co-planar then  $(\vec{A} \times \vec{B} ). \vec{C} = 0 \textrm{ or } \vec{A} ( \vec{B} \times \vec{C} = 0)$  

5.  $(\vec{A} \times \vec{B} ). \vec{C} = 0$   implies two of the vectors parallel.

6.  $(\vec{A} \times \vec{B} ). \vec{C} = \vec{A} .(\vec{B} \times \vec{C} ) = \vec{B}.(\vec{C} \times \vec{A} )$   = same volume.

    • Position vector:

  $$\begin{align*} & \textrm{OP } = x \hat{i} + y \hat{j} + z \hat{k} \\ & \textrm{PQ } = Q - P = ( x_1 - x ) \hat{i} + (y_1 - y) \hat{j} + (z_1 - z ) \hat{k} \\ & \textrm{OQ } = x_1 \hat{i} + y_1 \hat{j} + z_1 \hat{k} \end{align*}$$  

    Example: A plane is defined by any three points that are in the plane. If a plane contains the point  $P = (1,0,0), Q(1,1,1) \textrm{ and } R(2,- 1,3)$   then find a vector that is orthogonal to the plane.

 Solution: $$\begin{align*} & \vec{PQ} = \hat{j} + \hat{k} \\ & \vec{PR} = \hat{i} - \hat{j} + 3 \hat{k} \\ & | \vec{PQ} \times \vec{PR} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 1 & - 1 & 3 \end{array} \right| \end{align*}$$  

    Example: Find the area of the parallelogram with vectors  $P(2,2,0), Q(9,2,0), R(10,3,0) \textrm{ and } S(3,3,0 )$  

Solution:    $$\begin{align*} & \vec{QP}= 7 \hat{i} \\ & \vec{QR} = \hat{i} + \hat{j} \\ & |\vec{QP} \times \vec{QR} = \left| \begin{array}[]{ccc} \hat{i} & \hat{j} & \hat{k} \\ 7 & 0 & 0 \\ 1 & 1 & 0 \end{array} \right| & = |7 \hat{k} | \\ & = 7 \textrm{sqr unit} \end{align*}$$  

    Example: Find the area of the triangle with vectors  $A(1,0,0), B(0, 1,0), C(0,0,1 )$ .

Solution:  $$\begin{align*} \triangle ABC & = \frac{1}{2} | \vec{AB} \times \vec{AC} \\ & = \frac{1}{2} \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{array} \right| \\ & = \frac{1}{2} |\hat{i} + \hat{j} + \hat{k} | \\ & = \frac{1}{2} \sqrt{1^2 + 1^2 + 1^2} \\ & = \frac{\sqrt{3}}{2} \textrm{ sqr unit} \end{align*}$$  

    Example: find the volume of the parallelepiped, determined by the vectors  $(1,1,1), (4,7,2), (3,2,1), \textrm{ and } (5,4,3)$  

    Example: Determine, whether the three vectors  $\vec{A} = \hat{i} + 4 \hat{j} - 7 \hat{k} , \vec{B} = 2 \hat{i} - \hat{j} + 4 \hat{k}, \textrm{ and } \vec{C} = - 9 \hat{j} + 18 \hat{k}$   lie in the same plane or not.

Solution:  $$\begin{align*} \vec{C} (\vec{A} \times \vec{B}) & = \left| \begin{array}[]{ccc} 0 & - 9 & 18 \\ 1 & 4 & - 7 \\ 2 & - 1 & 4 \end{array} \right| \\ & = 9(4 + 14 ) + 18 (- 1 - 8) \\ & = 0 \end{align*}$$

$\therefore$ these vectors exists in the same plane.