IIT JAM

follow

Top Resources

Snehasish Ghosh posted an Question

July 08, 2020 • 05:21 am

30 points

IIT JAM
Physics (PH)

Please explain why is x taken as negative here..in the pic.

Please explain why is x taken as negative here in the pic.

2 Answer(s)

0 Likes
7 Comments
0 Shares

>
Deepak singh 1

it might be helpful ,
- Likes(0)
- Reply(0)

>
Dhairya sharma

If two points on a line are (x1,y1,z1) and (x2,y2,z2) then the line can be represented in vector form as (x2-x1)i^+(y2-y1)j^+(z2-z1)k^. Now in above question, for vector 'A' , the two known points are (1,0,0) and (0,2,0). Hence vector 'A' can be written as (0-1)i^+(2-0)j^+(0-0)k^ -i^+2j^. Similarly for vector 'B' the two known points are (1,0,0) and (0,0,3) hence 'B' can be represented as (0-1)i^+(0-0)j^+(3-0)k^ -i^+3k^
- Likes(2)
- Reply(4)
https://youtu.be/mSRxq6gNBns
- Like(1)
it will help u
- Like(1)
👍
- Like(0)
👍
- Like(0)

>
Deepak singh 1 Best Answer

see attached
- Likes(1)
- Reply(14)
if any doubt please ask.
- Like(1)
how is the coordinates of A (1,0,0)
- Like(1)
look at end points of vector A , see attached
- Like(0)
did not understand sir
- Like(1)
see this ,
- Like(1)
Sir,I will accept your answer..
- Like(1)
Sir if the points of A are on x axis and y axis,then why is the coordinate of A 1,0,0 ...I mean why is the y coordinate 0 then
- Like(0)
end points of vector A (point ) is on x axis... point
- Like(0)
It is also in y- axis na?
- Like(0)
points on x axis is of type ( x,0,0)
- Like(0)
But the vector A has end points in y axis also
- Like(0)
wait
- Like(0)
Sir,I understood
- Like(0)
I have used this result
- Like(1)

>
Chandra prakash

hope you get it dear
- Likes(0)
- Reply(0)

>
Chandra prakash

https://youtu.be/mSRxq6gNBns
- Likes(1)
- Reply(2)
go through this link...
- Like(0)
see it
- Like(0)

>
Chandra prakash

wait dear I want to send you link which is usefull for you
- Likes(1)
- Reply(0)

>
Ravi prakash yadav

If two points on a line are (x1,y1,z1) and (x2,y2,z2) then the line can be represented in vector form as (x2-x1)i^+(y2-y1)j^+(z2-z1)k^. Now in above question, for vector 'A' , the two known points are (1,0,0) and (0,2,0). Hence vector 'A' can be written as (0-1)i^+(2-0)j^+(0-0)k^ -i^+2j^. Similarly for vector 'B' the two known points are (1,0,0) and (0,0,3) hence 'B' can be represented as (0-1)i^+(0-0)j^+(3-0)k^ -i^+3k^
- Likes(0)
- Reply(2)
how is the coordinates of A (1,0,0)
- Like(0)
'A' is not a point,it is a vector. It has two end points having coordinates (1,0,0) and (0,2,0)
- Like(0)

T