Multiplication of Vector , Scalar Product , Vector Product

Multiplication Of a Vector By a Real Number :

The multiplication of a vector $\vec{A}$   by a real number k becomes another vector $ k \vec{A}$. its magnitude becomes k times the magnitude of the given vector. Its direction is the same or opposite as that of $\vec{A}$ ,  according as k is a positive real number or not.

Thus, $\displaystyle n (\vec{A}) = n \vec{A}$

And , $\displaystyle  – n (\vec{A}) = – (n \vec{A}) $

Multiplication Of a Vector By a Scalar:

When a vector $\vec{A}$  is multiplied by a scalar S, it become a vector $ S \vec{A}$ , whose magnitude is S times the magnitude of $\vec{A}$ and it acts along the direction of $\vec{A}$ . the unit of $ S \vec{A}$ , is different from the unit of vector $\vec{A}$ .
For illustration, if $\vec{A} = 100 $ newton due west and S = 10 sec, then

$ S \vec{A} = 10 \;second \times 100 \; Newton \; due \; west$  = 1000 Newton-second due west.

Multiplication Of A Vector By A Vector

Dot Product or Scalar Product :

The scalar product or dot product of two vectors $\vec{A}$ and $\vec{B}$ , written as $\vec{A}.\vec{B}$ (and read as $\vec{A}$ dot $\vec{B}$ ) is defined as the product of their magnitudes and cosine of angle between them,

i.e., $\vec{A} . \vec{B} = A B cos\theta $

The angle between two vectors is determined within the limits 0 to π when the vectors are made co-initial, i.e., their initial point coincide.

$\displaystyle \vec{A} . \vec{B} = \vec{B} . \vec{A}$  (commutative)

Self dot product of a vector gives square of magnitude of the vector, i.e., $\vec{A} . \vec{A} = A^2$

$\large \hat{i}.\hat{i} = \hat{j}.\hat{j} = \hat{k}.\hat{k} = 1$

$\large \hat{i}.\hat{j} = \hat{j}.\hat{k} = \hat{k}.\hat{i} = 0 $

Dot product of two vectors in terms of components

Let , $\vec{A}= A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and $\vec{B}= B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$

$\displaystyle \vec{A}.\vec{B} = (A_x \hat{i} + A_y \hat{j} + A_z \hat{k}).(B_x \hat{i} + B_y \hat{j} + B_z \hat{k})$

$\displaystyle = A_x B_x + A_y B_y + A_z B_z $

Cross Product or Vector Product :

The vector product or cross product of two vectors $\vec{A}$ and $\vec{B}$ , is written as $\vec{A} \times \vec{B}$ and is read as $\vec{A}$ cross $\vec{B}$ and is defined as a vector $\vec{C}$ whose product equal to the product of magnitudes of $\vec{A}$ and $\vec{B}$ and the sine of the angle between them,

the vector $\vec{C} = \vec{A} \times \vec{B} $ has a direction perpendicular to the plane of the two vectors $\vec{A}$ and $\vec{B}$ .

Thus , $ |\vec{A} \times \vec{B} = |\vec{A}| |\vec{B}| sin\theta \; or \; A B sin\theta $

The direction of vector $\vec{C} = \vec{A} \times \vec{B} $ is determined by the right hand screw rule (or right hand palm rule).

Also note that $\vec{A} \times \vec{B} = -\vec{B} \times \vec{A} $ , i.e., vector product of two vectors is not commutative

A unit vector is one whose magnitude equals unity. Any vector can be represented as the product of magnitude and a unit vector.

For example $\vec{A} = A \hat{A} $ where $\hat{A}$ is unit vector whose direction coincides with that of $\vec{A}$ .

$\displaystyle \hat{A} = \frac{\vec{A}}{A} $

$\displaystyle \hat{i} \times \hat{i} = \hat{j}\times\hat{j} = \hat{k}\times\hat{k} = 0 $

$\displaystyle \hat{i}\times\hat{j} = \hat{k} , \hat{j}\times\hat{k} = \hat{i} , \hat{k} \times \hat{i} = \hat{j} $

Vector Product or Cross product of two vectors in terms of components

Let , $\vec{A}= A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and $\vec{B}= B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$

$\displaystyle \vec{A} \times \vec{B} = (A_x \hat{i} + A_y \hat{j} + A_z \hat{k})\times (B_x \hat{i} + B_y \hat{j} + B_z \hat{k})$

Determinant Form :

$\displaystyle \vec{A} \times \vec{B} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{array} \right| $

Note:

The area of a triangle with side $\vec{A}$ and $\vec{B}$ is given by half of the magnitude of $ (\vec{A} \times \vec{B}) $

Area of triangle $\displaystyle = \frac{1}{2} |\vec{A} \times \vec{B}| $

The area of a parallelogram with sides $\vec{A}$ and $\vec{B}$ is given by the magnitude of $\vec{A} \times \vec{B}$

Area of parallelogram $\displaystyle = |\vec{A} \times \vec{B}| $

Projection/Component Of One Vector On Another Vector

Component of $\vec{A}$ along $\vec{B}$ = A cos θ

$\displaystyle = \frac{\vec{A} . \vec{B}}{B}$

Similarly , Component of $\vec{B}$ along $\vec{A}$ = B cos θ

$\displaystyle = \frac{\vec{A} . \vec{B}}{A}$

Projection of $\vec{B}$ on $\vec{A}$ is $ (B cos\theta )\hat{A} $ and vice versa .

Magnitude of projection of $\vec{B}$ on $\vec{A}$ is (B cosθ) and vice versa .

Position Vector And Change In Position

Position vector of a point (x , y , z) in space with respect to origin is given as

$ \vec{r} = (x \hat{i} + y \hat{j} + z \hat{k} ) $

If a particle moves from A (x1 , y1 , z1) to a point B(x2 , y2 , z2) then displacement i.e AB is given as

AB = position vector of B – Position vector of A

$ \vec{AB} = \vec{r_2} – \vec{r_1} $

$ \vec{AB} = (x_2 – x_1)\hat{i} + (y_2 – y_1)\hat{j} + (z_2 – z_1) \hat{k} $

Problem : If the two vectors $ \vec{A} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k} ) $ and $ \vec{B} = (\hat{i} + 2 \hat{j} – n \hat{k} ) $ are perpendicular then find the value of n .

Solution: For perpendicular vector , $ \vec{A} . \vec{B} = 0 $

∴ $ (2 \hat{i} + 3 \hat{j} + 4 \hat{k} ) . (\hat{i} + 2 \hat{j} – n \hat{k} ) = 0 $

2 + 6 – 4n = 0

or , -4 n = -8

or, n = 2

Problem  : What is the value of linear velocity, if $ \vec{r} = (3 \hat{i} – 4 \hat{j} + \hat{k} ) $ and $ \vec{\omega} = (5 \hat{i} – 6 \hat{j} + 6 \hat{k} ) $ ̂?

Solution : $\displaystyle \vec{v} = \vec{\omega} \times \vec{r} $

$\displaystyle \vec{v} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 5 & -6 & 6 \\ 3 & -4 & 1 \end{array} \right| $

Expanding along 1st row ,

$\displaystyle \vec{v} = \hat{i}(-6+24) -\hat{j}(5-18) + \hat{k}(-20+18)$

$\displaystyle \vec{v} = 18 \hat{i} + 13\hat{j} + 2 \hat{k}$

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