IIT JEE , AIEEE Forum - Mathematics , Vectors

srujana

A = 2i + 3 j

B= i + j

The component of vector A perpendicular to vector B and in the same plane is??

abhi_iit000

i think that we can get in this way..........formula is [AB]/[B]
so, [AB] is root13 and [B]is root2 so we get result as root of 13/2 ..................pls rate me if i'm not wrong...........

srujana

Options are :

1/

2 (j-i)

3/

2 (j-i)

5/

2 (j-i)

7/

2 (j-i)

a5hw1n_5

unit vector perpendicular to i+j is given by 1/

2(j-i) [or i-j ]

therefore mag of component of A is given by (2i+3j).1/

2(j-i) =(3-2)/

2=1/

2.

therefore the vector is 1/

2(1/

2(j-i)) = 1/2(j-i)

hence option a is right

dhinakaran

is tht not zero?..cos compenent of A prpendicular to B....so magnitude of Acos90 ...so its zero.....im confused plzz tell me which one correct

a5hw1n_5

u read the whole que wrong, the vector is perpendicular to B and not to A.

luxmi

If angle btwn A and B is @. vector component of A prpndicular to B is Asin@.

frm cross product rule

ABsin@ = |AXB|

Asin@ = |AXB| / |B| [this is for the magnitude of the component only,

direction is either (j-i) or (i-j)]

AXB = +k or -k

|AXB|= 1

|B| =

2

so Asin@= 1/

2(j-i) option (A)

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