Q-1
The projection or component of #vecA# along #vecB# is #(vecA*vecB)/absvecB^2vecB#
So the component of #veca=3hati+4hatj# along the direction of #hati+hatj# will be
#((3hati+4hatj)*(hati+hatj))/abs(hati+hatj)^2(hati+hatj)#
#=(3+4)/(sqrt2)^2(hati+hatj)#
#=7/2(hati+hatj)#
And the component of #veca=3hati+4hatj# along the direction of #hati-hatj# will be
#((3hati+4hatj)*(hati-hatj))/abs(hati-hatj)^2(hati-hatj)#
#=(3-4)/(sqrt2)^2(hati-hatj)#
#=-1/2(hati-hatj)#
Q-2
Let #(x,y) # be the cartesian coordinates of the point A represeted by the radius vector #vecr=athati-bt^2hatj#
So #x=at and y=-bt^2#
Eliminating t from above relation we get
#y=-b(x/a)^2#
#=>y=-b/a^2x^2#
#=>x^2=-a^2/by#
This is the equation of the trajectory represented by the radius vector #vecr=athati-bt^2hatj#
