Hello fellow humans,
Again I am pestering you about the Cambridge text, just to make sure I am not going crazy (though the quest for the meaning of pi probably does that to everyone mathematically inclined).
If someone doesnt mind, could this special person please tell me how to write in mathematical text on here? I see a "pi" symbol that highlights as "LaTeX", but I am not very savvy when it comes to computer coding.....
I have attached the question concerned, so I will run through my effort thus far:
Part (a)
(i) Just asks for vector OD. If AD = 1/2AB, then OD = 1/2AB + OA = 1/2(b+a)
(ii) Asks for DE, so that will be OE - OD and since OE = kOB, then
DE = kb - 1/2(b+a)
= (k-1/2)b -1/2a
Part (b)
Showing the value of lamda (k) given if DE is perpendicular to OB just requires the dot product of the two vectors being zero
DE.OB = [(k-1/2)b -1/2a].b
= (k-1/2)b.b -1/2a.b
= kb.b -1/2b.b -1/2a.b = 0
kb.b = 1/2(a.b + b.b)
k = (a.b + b.b)/(2b.b)
Part (c)
(i) Finding the value of the cosine just requires fiddling with the expression for lamda in (b)
5/6 = (a.b + b.b)/(2b.b)
5/3 = a.b/b.b +1
2/3 = a.b/b.b
If a.b = (a)(b)cosx, with x the angle AOB, then
(a)(b)cosx/(b)^2 = 2/3
(a)/(b)cosx = 2/3
As OA = OB, then (a) = (b), thus
cosx = 2/3 Q.E.D.
(ii) This final part is the most excruciating, though it shouldnt be if I had added the vectors correctly. Please tell me I am not going crazy?
Firstly OD = 1/2(b+a)
DE = (k-1/2)b -1/2a, and since k = 5/6
= 1/3b - 1/2a
Then DF = 1/2DE = 1/6b - 1/4a
DF = OF - OD, thus
OF = DF + OD
= 1/6b - 1/4a + 1/2a + 1/2b
= 2/3b + 1/4a
AE = OE - OA = 5/6b - a
If AE and OF are perpendicular, then their dot product SHOULD be zero
AE.OF = [2/3b + 1/4a].[5/6b - a] = 10/18(b)^2 - 2/3a.b + 5/24a.b - 1/4(a)^2
Taking the fact (b) = (a) and (b)^2 out as a common factor as a result
AE.OF = (b)^2[10/18 - 2/3 + 5/24 - 1/4]
Making 72 the common denominator
AE.OF = (b)^2[40/72 - 48/72 + 15/72 - 18/72] = -11/72(b)^2
The above is obviously not zero, and I have been through the calculation a number of times to make sure I did not miss anything.
Give it a go. I hope I am wrong.
James