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Establish the vector inequalities geometrically or otherwise :

| a - b | ≥ || a | - | b ||

Let two vectors a and b be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.


The following relations can be written for the given parallelogram. 

OS + PS > OP                               ...(i) 

OS > OP - PS                                ...(ii) 

| a - b | > | a | - | b |                    ...(iii) 

The quantity on the LHS is always positive and that on the RHS can be positive or negative.

To make both quantities positive, we take modulus on both sides.

That is, 

| | a - b | | > | | a | - | b | | 

| a - b | > | | a | - | b | |                 ...(iv) 

If the two vectors act in a straight line but in the opposite directions, then we can write,

| a - b | = | | a | - | b | |                   ...(v)

Combining equations (iv) and (v), we get, 

| a - b | ≥ | | a | - | b | |  
 
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Establish the vector inequalities geometrically or otherwise :
(c) |ab| < |a| + |b|



Let two vectors a and b be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.


 

Here we have: 

| OR | = | PS | = | b |              ...(i) 

| OP | = | a |                          ...(ii) 

In a triangle, each side is smaller than the sum of the other two sides. Therefore, in ΔOPS, we have: 

OS < OP + PS 

| a - b | < | a | + | -b | 

| a - b | < | a | + | b |            ... (iii) 

If the two vectors act in a straight line but in opposite directions, then we have,

| a - b | = | a | + | b |             ... (iv) 

Combining equations (iii) and (iv), we get, 

| a - b | ≤ | a | + | b | 
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A man can swim with a speed of 4.0 km/hr in still water. How long does he take to cross the river 1.0 km wide if the river flows steadily at 3.0 km/hr and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?

Since the swimmer dives in the river normal to the flow of river, therefore time taken by swimmer to cross the river is,
space space space space space space space straight t equals straight d over straight v subscript straight m equals fraction numerator 1 km over denominator 4 km divided by hr end fraction equals 1 fourth hr equals 15 space min

Since the swimmer dives in the river normal to the flow of river, the
In this time the swimmer will also go down the river by distance BC due to river current.
∴   BC space equals space straight v subscript straight r straight t space equals space left parenthesis 3 km divided by hr right parenthesis open parentheses 1 fourth hr close parentheses
            equals 3 over 4 km equals 0.75 km

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A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest taxi driver takes him along a circuitous path 23 km long and reaches the hotel in 28 minutes.
What is:
(a) the average speed of the taxi?
(b) the magnitude of average velocity? Are the two equal? 


Given,

Magnitude of displacement = 10 km

Total path length = 23 km

Time taken = 28 min = 7/15 hr.

Therefore,



                        

Magnitude of average velocity is,

      

Therefore, we can see that the average speed and magnitude of average velocity are not equal.

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A cyclist starts from the center O of circle of park of radius 1 km, reaches the edge P of the park, then cycles along the circumference to point Q and returns to the centre along QO as shown. If the round trip takes 10 minutes, then what is:

(a) net displacement
(b) average velocity
(c) average speed of the cyclist?

(a) The cyclist starts from O and again returns to point O. Therefore, the initial and the final point are the same. Hence, displacement of cyclist is equal to zero.

(b) Average velocity of cyclist is zero because total displacement of cyclist is zero.

(c) Total distance travelled by cyclist is,

             

                

∴      Average speed  = 

and

Time taken for round trip,

∴   

                         

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