## Note on Vector Subtraction and Multiplication

• Note
• Things to remember

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

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
1. It does not follow commutative law. $$\vec A \times \vec B \neq \vec B \times\vec A$$ but, \, $$\vec A \times \vec B = -\vec B \times\vec A$$
2. It follows distributive law. $$\vec A \times (\vec B + \vec C) = \vec A \times \vec B + \vec A \times \vec C$$
3. 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.

Vector subtraction is defined as the addition of one vector to the negative of another.

The process of splitting the single vector into many components is called the resolution of vectors.

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

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

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

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##### makshye narayan sainju

must above mathematics compulsory recall for exam?
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