Author Topic: Vector Proof... (Read 1465 times) Tweet Share

nerd · « **on:** February 13, 2009, 07:46:34 pm »

I'm having trouble with Question 20 in exercise 2E in Cambridge Essential Specialist Maths. Can anyone help? (Sorry for those without the textbook, I cbf writing out the whole question because it is quite a long one!)

Thanks.

Edmund · « **Reply #1 on:** February 13, 2009, 08:04:38 pm »

Mao · « **Reply #2 on:** February 13, 2009, 08:29:18 pm »

I remember doing this question, though not how I actually did it. So here goes nothing:

$EX = k_1(EB + BC)$

$FX = k_2 (FA + AD)= k_2 (-EB + BC)$

$EX\cdot FX = k_1k_2 (EB+BC)(-EB+BC) = k_1k_2 (BC\cdot BC - EB\cdot EB ) = k_1k_2 (|BC|^2 - |EB|^2) = 0$ (as they are the same length)

$\therefore EX\perp FX$ QED

Mao · « **Reply #3 on:** February 13, 2009, 09:06:55 pm »

b)

$FE = FX + XE$

$(2+k)BE = \mu FD + \lambda CE$

$(2+k) BE = \mu (FA + AD) + \lambda (CB + BE)$

$(2+k) BE = \mu (BE + BC) + \lambda (BE - BC)$

$(2+k) BE = (\mu + \lambda) BE + (\mu - \lambda)BC$

since BE and BC are linearly independant [not parallel], we equate the coefficients:

$2+k = \mu + \lambda,\; \mu - \lambda = 0$

$\implies \mu = \lambda = \frac{k+2}{2}$

c)

$k=1\; \implies \; \mu = \lambda = \frac{3}{2}$

Mao · « **Reply #4 on:** February 13, 2009, 09:21:53 pm »

d)

$|FX| = |EX|$

$\mu |FD| = \lambda |EC|$

$|FD| = |EC|$

$|FA + AD| = |EB + BC|$

$|BE + BC| = |EB + BC|$

$|BE + BC|^2 = |EB + BC|^2$

$(BE + BC)\cdot (BE + BC) = (-BE + BC) \cdot (-BE + BC)$

$|BE|^2 + 2BE\cdot BC + |BC|^2 = |BE|^2 - 2BE\cdot BC + |BC|^2$

$4BE\cdot BC = 0$

$BE\cdot BC = 0$

$\implies BE \perp BC$

$\implies AB \perp BC$

$\therefore \mbox{ABCD is a rectangle}$ QED

Mao · « **Reply #5 on:** February 13, 2009, 09:34:01 pm »

yah, I remember having much trouble when I first did this question... Took me a couple nights before I got it done

/0 · « **Reply #6 on:** February 13, 2009, 10:02:56 pm »

Lol I remember using the cosine rule to prove each angle was 90 degrees and so proving it was a rectangle

TrueTears · « **Reply #7 on:** February 14, 2009, 12:27:40 pm »

lol banksy just did this Q for me

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Author Topic: Vector Proof... (Read 1465 times) Tweet Share

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Vector Proof...

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