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Topics Covered

• Multiplication of Vectors by Real Numbers
• Addition & Subtraction of Vectors - Graphical Method
• Head-to-Tail Method
• Triangle Method
• Parallelogram Method

Multiplication of Vectors by Real Numbers

`text(Multiplication with a Positive Number)`
Multiplying a vector `vecA` with a positive number `lamda` gives a vector whose magnitude is changed by the factor `lamda` but the direction is the same as that of `vecA`.
`|lamdavecA|=lamda|vecA|` if `lamda > 0`

For example, if `vecA` is multiplied by 2, the resultant vector `2vecA` is in the same direction as `vecA` and has a magnitude twice of `|vecA|` as shown in Fig. (a).

`text(Multiplication with a Negative Number)`
Multiplying a vector `vecA` by a negative number `lamda` gives a vector `lamdavecA` whose direction is opposite to the direction of `vecA` and whose magnitude is `–lamda` times `|vecA|`.

Multiplying a given vector `vecA` by negative numbers, say –1 and –1.5, gives vectors as shown in Fig (b).

Addition & Subtraction of Vectors - Graphical Method

Graphical methods of vectors addition and subtraction -
• Head-to-Tail Method
• Triangle Method
• Parallelogram Method

Head-to-Tail or Triangle Method of Vector Addition

Let us consider two vectors `vecA` and `vecB` that lie in a plane as shown in Fig. (a). The lengths of the line segments representing these vectors are proportional to the magnitude of the vectors.
To find the sum `vecA + vecB`, we place vector `vecB` so that its tail is at the head of the vector `vecA`, as in Fig. (b). Then, we join the tail of `vecA` to the head of `vecB`. This line OQ represents a vector `vecR`, that is, the sum of the vectors `vecA` and `vecB`. Since, in this procedure of vector addition, vectors are arranged head to tail, this graphical method is called the head-to-tail method.

The two vectors and their resultant form three sides of a triangle, so this method is also known as `text(triangle method of vector addition)`.

• If we find the resultant of `vecB + vecA` as in Fig. (c), the same vector `vecR` is obtained. Thus, vector addition is commutative.
`vecA + vecB = vecB + vecA`
• The addition of vectors also obeys the associative law as illustrated in Fig. (d). The result of adding vectors `vecA` and `vecB` first and then adding vector `vecC` is the same as the result of adding `vecB` and `vecC` first and then adding vector `vecA`.
`(vecA + vecB) + vecC = vecA + (vecB + vecC)`

`text(Null Vector or Zero Vector)`
Consider two vectors `vecA` and `–vecA`. Their sum is `vecA + (–vecA)`. Since the magnitudes of the two vectors are the same, but the directions are opposite, the resultant vector has zero magnitude and is represented by `vec0` called a `text(null vector)` or a `text(zero vector)`.
`vecA – vecA = vec0`
`|vec0|= 0`
Since the magnitude of a null vector is zero, its direction cannot be specified.

The null vector also results when we multiply a vector A by the number zero. The main properties of `vec0` are -
`vecA + vec0 = vecA`
`lamda vec0 = vec0`
`0 vecA = vec0`

`text(Subtraction of Vectors)`
Subtraction of vectors can be defined in terms of addition of vectors. We define the difference of two vectors `vecA` and `vecB` as the sum of two vectors `vecA` and `–vecB`.
`vecA – vecB = vecA + (–vecB)`
It is shown in Fig (e,f). The vector `–vecB` is added to vector `vecA` to get `vecR_2 = (vecA – vecB)`.

Parallelogram Method

Suppose we have two vectors `vecA` and `vecB`. To add these vectors, we bring their tails to a common origin O as shown in Fig. (a).
Then we draw a line from the head of `vecA` parallel to `vecB` and another line from the head of `vecB` parallel to `vecA` to complete a parallelogram OQSP. Now we join the point of the intersection of these two lines to the origin O.

The resultant vector `vecR` is directed from the common origin O along the diagonal (OS) of the parallelogram [Fig. (b)].

In Fig. (c), the triangle law is used to obtain the resultant of `vecA` and `vecB` and we see that the two methods yield the same result. Thus, the two methods are equivalent.
Q 3015778669

Rain is falling vertically with a speed of `35 m s^(–1)`. Winds starts blowing after sometime with a speed of `12 m s^(–1)` in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella ?


The velocity of the rain and the wind are represented by the vectors `v_r` and `v_w` in Fig. and are in the direction specified by the problem. Using the rule of vector addition, we see that the resultant of `v_r` and `v_w` is `R` as shown in the figure. The magnitude of `R` is

`R = sqrt(v_r^2+v_w^2) = sqrt((35)^2+(12)^2) m s^(-1) = 37 ms^(-1)`

The direction `θ` that `R` makes with the vertical is given by `tantheta = v_w/v_r = 12/35 = 0.343`

or `theta = tan^(-1) (0.343) = 19^0`

Therefore, the boy should hold his umbrella in the vertical plane at an angle of about `19^0` with the vertical towards the east.