# Vector Subtraction and Multiplication

## Subtraction of the vectors

Vector subtraction is defined as the addition of one vector to the negative of another. Subtraction of vector \(\vec A \; \text{from} \; \vec B\) means addition of vector \(- \vec A \; \text {with} \; \vec B\). We can analyze the subtraction of vector as shown in the figure. The resultant between \(\vec A \; \text{and} \; \vec B\) gives the resultant of the sum of vectors while the resultant between \(- \vec A \; \text {and} \; \vec B\) of difference of vectors.

$$|- \vec A| = \vec A = A$$

$$| \vec B| = \vec B= B$$

$$R =\sqrt {A^2 + 2AB\cos(180^o - \theta) + B^2} $$

$$R =\sqrt {A^2 - 2AB\cos\theta + B^2} $$

Similarly,

$$\beta=\tan^{-1}\frac{B\sin(180^o - \theta)}{(A + B\cos(180^o -\theta)}$$

$$=\tan^{-1}(\frac{B\sin \theta}{A - B\cos\theta})$$

Vector subtraction does not follow commutative and associative law.

## Resolution of vectors

The process of splitting the single vector into many components is called the resolution of vectors. A vector can have number of component vectors.

Suppose a vector \(\vec A\) shown in the figure. The component of \(\vec A\) along OX-direction is OQ. In POQ,

$$\cos\alpha = \frac {OQ}{OP} = \frac {OQ}{A}$$

$$OQ = A \cos\alpha$$

Similarly the component of along OX'-direction is\(OT = A\cos\beta\) where \(beta\)is angle made by with OX-direction and that along OX''-direction is \(OT = A\cos\alpha\)

**Rectangular Components of a vector**

When a vector is resolved \(\vec A\) into perpendicular directions, then the component vectors \(\vec A\)are called rectangular components of the vector.

Let us consider a vector \(\vec A\) represented by OB in x-y plane. Suppose that \(\vec A\) makes an \(\theta\) with x-axis. Let us drop perpendicular on x-axis and y-axis at P and Q. Then OP and OQ represents the components of \(\vec A\), such that \(\vec A =\vec A_x + \vec A_y \), where \(\vec A_x and \vec A_y\) represents the X component and the Y component of \(\vec A\) respectively.

$$ In \Delta OBP, $$

$$ \sin\theta = \frac{BP'}{OP} $$

$$BP =OB \sin\theta $$

$$A_y = Asin\theta \dots (i)$$

$$ \cos\theta = \frac{OP}{OB} $$

$$OP =OB \cos\theta $$

$$A_x = Acos\theta \dots (ii)$$

Squaring and adding equation (i) and (ii)

$$ A =\sqrt {A_X^2 + A_Y^2} $$

Dividing equation (i) and (ii)

$$\tan\theta = \frac{A_x}{A_y}$$

**Vector Multiplication**

**Scalar product:** It is defined as the multiplication of magnitude of one vector to the scalar component of another vector in the direction of the first vector.

Let two vectors \(\vec A and \vec B\) inclined at an angle to each other as shown in the figure.

$$\vec A . \vec B = AB\cos\theta$$

##### Properties of Scalar Product

- Dot product is always a scalar quantity and is positive if \(\theta \lt 90^o and 180^o \gt \theta \gt 90^o\)
- It follows commutative law i.e. \(\vec A . \vec B = \vec B . \vec A\)
- It follows distributive law i.e.\(\vec A . (\vec B + \vec C) = \vec A .\vec B + \vec A .\vec C \)
- (\(\vec A)^2 = \vec A.\vec A = A^2\)
- \(\vec A .\vec B)_{max.} = AB \text{when} \theta = 0^o\)
- \(\vec A .\vec B)_{min.} = 0 \text{when} \theta =9 0^o(A \neq 0, B = 0)\)

**Vector product:**

If the product of two vectors is a vector quantity, then such an operation is called vector product.

Let two vectors \(\vec A and \vec B\) inclined at an angle to each other as shown in the figure.

$$\vec A \times \vec B = AB\sin\theta \, \widehat n$$

Where \(\widehat n\) is a unit vector in the direction perpendicular to the plane of \(\vec A and \vec B\) and Torque due to a force is the cross product of force and perpendicular distance.

$$\vec \tau = \vec r \times \vec F$$

### Properties of cross-product

- It does not follow commutative law. \(\vec A \times \vec B \neq \vec B \times\vec A\) but, \, \(\vec A \times \vec B = -\vec B \times\vec A\)
- It follows distributive law. \(\vec A \times (\vec B + \vec C) = \vec A \times \vec B + \vec A \times \vec C \)
- If a vector is multiplied by itself the result becomes \( 0. \vec A \times \vec A = A^2=\sin 0^o = 0\)

**Geometrical interpretation of Cross Product**

Let us consider two vectors \(\vec A \,and \, \vec B\) represented in magnitude and direction by \(\vec OZ \,and \, \vec OX\). Let us complete the parallelogram OXYZ and drop a perpendicular from X at N.

$$\vec A \times \vec B = AB\sin\theta$$

$$ \text {Area of parallelogram OXYZ} = \text{(base)} \times\text{ (perpendicular distance)} $$

$$= (OZ) \times (XN) $$

$$= (OZ)(OX\sin\theta)$$

$$=(A)(B\sin\theta)$$

$$=AB\sin\theta$$

That is an area of the parallelogram is equal to the magnitude of the cross product of two vectors which are represented by two adjacent sides of a parallelogram.

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