Vectors...really really need help!?

With respect to the origin O, the position vectors of points A, B and C are (2, 3), (4, -2) and (-3, -3) respectively.





(i) A point E (-2, p) is such that 2 |CE| = |AB|. Find the possible values of p.


Ans: - 5 1/2 or -1/2





(ii) Find the coordinates of the point Q if 2 CQ = BA + 1/2 AC.


Ans: (-5 1/4, -2)





Please help by showing workings...thanks

Vectors...really really need help!?
(i)


|CE|^2 = (-3 - (-2))^2 + ((-3) - p)^2


= 1 + 9 + p^2 + 6p


= p^2 + 6p + 10.





|AB|^2 = (2 - 4)^2 + (3 - (-2))^2


= 4 + 25


= 29





Therefore, as 4 CE^2 = AB^2:


4p^2 + 24p + 40 = 29


4p^2 + 24p + 11 = 0


(2p + 11)(2p + 1) = 0


p = -11/2 or - 1/2.





(ii)


BA = (2, 3) - (4, -2) = (-2, 5)


AC = (-3, -3) - (2, 3) = (-5, -6)





2CQ = (-2 , 5) + (1/2)(-5, -6)


= (-2 - 5/2, 5 - 3)


= (- 9/2, 2) ...(1)





If Q is (x, y), then:


2CQ = 2(x - (-3), y - (-3))


= (2x + 6, 2y + 6) ...(2)





Equating components from (1) and (2):


2x + 6 = -9/2


x = ( - 9/2 - 6) / 2 = -21/4,


and


2y + 6 = 2


y = (2 - 6) / 2 = - 2





Thus Q = (-21/4, - 2).
Reply:A) 2i+3j


B)4i-2j


C)-3i-3j


AB ) 2i-5j and IABI =sqrt(29)


CE ) i+( p+3)j and ICE I = sqrt(1+(p+3)^2)


2sqrt((1+(p+3)^2) = sqrt29


4 +4(p+3)^2 =29


4p^2+24p+11=0 p= (-24+-20)/8 so p= -1/2 and p=-5 1/2


Call Q) x i+y j


2[(x+3)i+(y+3)j] = -2i+5j -5/2i-3j


2x+6= -9/2 x= -21/4 = -5 1/4


2y+6=2 so y=-2