Let `vecb= 4 hati + 3hatj and vecc` be two vectors perpendicular to each other in the xy- plane. All vectors in the sme plane having projections 1 and 2 along `vecb and vecc`., respectively, are given by ________

Answer

`(2hat(i) -hat(j))`

Explanation

Let `vec(c ) = a hat(j) + b hat(j)`
since `vec(b) " and " vec(c )` are perpendiculars to each other . Then
`vec(b) "." vec( c) =0 rArr (4hat(i) +3hat(j)) .(a hat(j) +b hat(j)) =0`
`rArr 4a + 3b=0 rArr a: b =3 :-4`
`:. , vec(c ) = lambda (3 hat(i) -4 hat(j)) ,` where `lambda` is constant of ratio. ,
Let the requried vectors be `vec(a) = p hat(i) + q hat(j)`
Projection of `vec(a) " on " vec(b) " is " (vec(a) ". " vec(b))/(|vec(b)|)`
`:. 1= (4p+3q)/(5) rArr 4p+ 3q=5`
Also projection of `vec(a) "on " vec(c ) " is " (vec(a) ". " vec(c ))/(|vec(c )|)`
`rArr 2= (3lambda p - 4lambda q)/(5lambda) rArr 3p -4q=10`
On solving above equations we get p=2 , q=-1
`:. vec(c ) =2hat(i) - hat(j)`