Question

ok here is problem that doesn't go into my mind hehe

Answer 1

the dot product definition \(\vec{a}.\vec{b}=||a|||b||cos\theta\)
theta angle between vectors a and b
some text books tend to explain this using work force problem
which doesn't look to me intuitive!
i doesn't seem to me they make any relation to cosine law

Answer 2

somehow cosine of angle between \(\large \vec{a}\) and \(\large \vec{b} \) is sneaking in \(\large \vec{a}\bullet \vec{b}\)

Answer 3

so basically they from work-force problem we can see that
a is projected on b and so...
and first of all how do i know that the force forms and angle with the direction of the object?

Answer 4

do you know how to swim ?

Answer 5

say* first line

Answer 6

yeah! sort of

Answer 7

is it easy to swim against the flow of water
or along the the flow water ?

Answer 8

easy to swim with the flow

Answer 9

water flow does some positive work when you're along its direction, yes ?

Answer 10

yes! that is logical

Answer 11

if you're going in reverse direction,
then water flow does negative work, yes ?

Answer 12

yes!

Answer 13

if you're going PERPENDICULAR to the water flow,
then water does 0 work, still yes ?

Answer 14

true

Answer 15

So the amount of work done by water flow depends on the angle between your swimming path and the flow path.

Answer 16

Suppose water flow = \(\large \overrightarrow{W}\)
your swimming direction = \(\large \overrightarrow{S}\)

Answer 17

hmmm sounds good

Answer 18

then the work done can be represented using the dot product formula :
\[\large \text{work done by water in moving you} = \overrightarrow{W} \bullet \overrightarrow{S}\]

Answer 19

which equals \[\large \overrightarrow{W} \bullet \overrightarrow{S} = |\overrightarrow{W} | |\overrightarrow{S}| \cos \theta \]

because for obvious reasons

Answer 20

i see... so we know it is cosine
because of cosine law? right?

Answer 21

the work done must become 0 when, \(\theta = 90\)
and it must be positive and decrease as \(\theta \) increases from \(\large 0 \to 90\)

Answer 22

yes, there is a nice way for representing dot product in cartesian coordinates also :

\(\large \langle a_1, a_2, a_3 \rangle \bullet \langle b_1, b_2, b_3\rangle = a_1b_1 + a_2b_2 + a_3b_3 \)

Answer 23

\[\overrightarrow{W} \bullet \overrightarrow{S} = |\overrightarrow{W} | |\overrightarrow{S}| \cos \theta = \langle a_1, a_2, a_3 \rangle \bullet \langle b_1, b_2, b_3\rangle = a_1b_1 + a_2b_2 + a_3b_3 \]

Answer 24

yeah i know of that! that one has no cosine involved at all which is a mystery hehehe

Answer 25

cosine is still hiding in those coordinates :)

Answer 26

and it must be positive and decrease as θ increases from 0→90
hold one what are saying here in this line?
the work can be negative as well no?
what about the angle can as well

Answer 27

yes

Answer 28

oh i see so it is simplified!

Answer 29

the work done by "water flow" in moving you can be positive/negative/0

Answer 30

depends on the angle, is that right!

Answer 31

when you're falliing down, gravity is doing positive work
when you're jumping up, gravity is doing negative work

Answer 32

yes, cosine is negative in second and third quadrants

Answer 33

negative work is same as
swimming against the water flow

Answer 34

the water flow is working against your movement

Answer 35

yes!
i still don't get what you said about decreases part? can you elaborate
are you saying that work decrease as the angle get bigger!

Answer 36

oh yeah i got it as angle goes to 90 the work decreases
thank @ganeshie8