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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: bts on April 25, 2014, 05:29:05 pm

Title: Maths Quest!!!!
Post by: bts on April 25, 2014, 05:29:05 pm: Question 5ci) is confusing it was on the VCAA 2003 exam 2 but i still don't get it

HELP????
Title: Re: Maths Quest!!!!
Post by: Conic on April 25, 2014, 08:45:27 pm: Let $\underset{\sim}{u}=p\underset{\sim}{i}+q\underset{\sim}{j}+r\underset{\sim}{k}$ . We know $\overrightarrow{OA} \cdot \underset{\sim}{u}=\overrightarrow{OC}\cdot \underset{\sim}{u}=0$ , since u is perpendicular to OA and OC. From this we get 2 equations:

$-2p+5q-2r=0$ (From dot product of u and OA)

$4p+q+4r=0$ (From dot product of u and OC)

Combining these gives

$4p+q +4r+2(-2r+5q-2r)=0-2\cdot0$

$\implies q+10q=0$

$\implies q=0$

So that shows why q=0. Now the equations become:

$-2p-2r=0$

$4p+4r=0$

So we have p=-r. Since u is a unit vector, we know that the magnitude is 1, so

$p^2+r^2=1$

$\implies p^2+(-p)^2=1$ (since r=-p)

$\implies 2p^2=1$

$\implies p=\frac{1}{\sqrt{2}}$ (since p>0)

$\implies r=-\frac{1}{\sqrt{2}}$ (since r=-p)

So $\underset{\sim}{u}=\frac{1}{\sqrt{2}}\underset{\sim}{i}-\frac{1}{\sqrt{2}}\underset{\sim}{k}$
Title: Re: Maths Quest!!!!
Post by: bts on April 25, 2014, 09:02:34 pm: thank you so so much :)

A massive help...

but for the next part isn't it just r cause thats the k component which is the height?
Title: Re: Maths Quest!!!!
Post by: Conic on April 25, 2014, 09:09:02 pm: The height of the pyramid is the length of DE, which is not necessarily the k component.
Title: Re: Maths Quest!!!!
Post by: bts on April 25, 2014, 09:38:29 pm: would you assume E is half OB cause otherwise not sure how to do it :(
Title: Re: Maths Quest!!!!
Post by: Conic on April 25, 2014, 09:44:19 pm: D is not necessarily over the centre of the base, so you need to use the information from the previous question.

DE is the vector resolute of DO in the direction perpendicular to the base of the pyramid (try to make sense of this by looking at the diagram). In the last part of the question we found the unit vector in that direction (the vector we found was perpendicular to OA and OC, so it is perpendicular to the base), so we can find DE:

$\overrightarrow{DE}=(\overrightarrow{DO}\cdot \underset{\sim}{u})\underset{\sim}{u}$

$=((-3\underset{\sim}{i}-4\underset{\sim}{j}+\frac{1}{3}\underset{\sim}{k})\cdot(\frac{1}{\sqrt{2}}\underset{\sim}{i}-\frac{1}{\sqrt{2}}\underset{\sim}{k}))\underset{\sim}{u}$

$=(\frac{-3}{\sqrt{2}}-\frac{1}{3\sqrt{2}})\underset{\sim}{u}$

$=-\frac{10}{3\sqrt{2}}\underset{\sim}{u}$

So DE has a length of $\frac{10}{3\sqrt{2}}$ .
Title: Re: Maths Quest!!!!
Post by: bts on April 25, 2014, 09:46:32 pm: the ans says its 5/(3squareroot2) though?
Title: Re: Maths Quest!!!!
Post by: Conic on April 25, 2014, 09:53:39 pm: The VCAA assessment report says its $\frac{10}{3\sqrt{2}}$ or $\frac{5\sqrt{2}}{3}$ . I'd take these over maths quest, as textbooks do have errors sometimes.
Title: Re: Maths Quest!!!!
Post by: bts on April 25, 2014, 09:57:48 pm: oh ok thank you
can you please exaplain this bit:
we found the unit vector in that direction (the vector we found was perpendicular to OA and OC) thats in the direction of the base? how did you know?
Title: Re: Maths Quest!!!!
Post by: Conic on April 25, 2014, 10:02:57 pm: We were given that the unit vector we were looking for was perpendicular to OA and OC. OA and OC make up the base of the pyramid, so a vector perpendicular to OA and OC is perpendicular to the base. They also tell us that E is the point on the base such that DE is perpendicular to the base. Therefore know that both DE and the unit vector are perpendicular to the base. This means the unit vector and DE are parallel, because they are in the same direction.
Title: Re: Maths Quest!!!!
Post by: bts on April 25, 2014, 10:10:17 pm: oh thank you

how did you get from

Quote from: Conic on April 25, 2014, 09:44:19 pm

$=-\frac{10}{3\sqrt{2}}\underset{\sim}{u}$

So DE has a length of $\frac{10}{3\sqrt{2}}$ .
Title: Re: Maths Quest!!!!
Post by: Conic on April 25, 2014, 10:13:33 pm: The length of u is 1, so if you multiply it by $-\frac{10}{3\sqrt{2}}$ its new length will be $\left |1\times -\frac{10}{3\sqrt{2}} \right |=\frac{10}{3\sqrt{2}}$ .
Title: Re: Maths Quest!!!!
Post by: bts on April 25, 2014, 10:22:26 pm: thank you so so much :
just curious: how did they get the (5squareroot2) / 3
Title: Re: Maths Quest!!!!
Post by: Conic on April 25, 2014, 11:05:35 pm: By rationalising the denominator:

$\frac{10}{3\sqrt{2}}=\frac{10}{3\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{10\sqrt{2}}{3\times 2}=\frac{5\sqrt{2}}{3}$
Title: Re: Maths Quest!!!!
Post by: bts on April 25, 2014, 11:09:26 pm: OMG im so stupid :)

Thank you

respect +1 thank you, have you looked at the question from 1MIN? i tried doing it and then stopped half way... pretty sure you'll be able to ace that :)
Title: Re: Maths Quest!!!!
Post by: Conic on April 25, 2014, 11:22:15 pm: They answered that question in another thread recently: it's so hard :(
Title: Re: Maths Quest!!!!
Post by: bts on April 25, 2014, 11:34:41 pm: yea... i know... but i still couldn't ans it cause before P lies on a straight line so you knew one of the coordinates already being 7, but with this one the P is on a slanting line so i'm not sure about any of the co ordinates making it hard... maybe its just me?????
Title: Re: Maths Quest!!!!
Post by: Conic on April 26, 2014, 01:48:11 pm: So you let P=(x,y). Now

$\overrightarrow{ZP} = \overrightarrow{OP} - \overrightarrow{OZ} = x \underset{\sim}{i} + y \underset{\sim}{j} -5\underset{\sim}{i} = (x-5)\underset{\sim}{i}+y \underset{\sim}{j}$

We know that $\overrightarrow{OY} = \overrightarrow{OX} + \overrightarrow{OZ} = 7 \underset{\sim}{i} + 7 \underset{\sim}{j}$ , and we know that ZP and OY are perpendicular, so
$\overrightarrow{ZP} \cdot \overrightarrow{OY} = 0$

$\implies 7\times (x-5)+7 \times y = 0$

$\implies x = 5-y$

We also know that y=7, so x=5-7=-2.
Title: Re: Maths Quest!!!!
Post by: bts on April 26, 2014, 02:10:56 pm: but i think the ans says (12 and 3/8, 10 and 5/8)

i think the x = -2 is for the other question:

What i did was:
Let P (xi + yj)
XP . XZ = 0
XP = (-3+x)i +(-5+y)j
XZ = (3i - 5j)
[(-3+x)i +(-5+y)j] . (3i - 5j) = 0
-9 + 3x +25 -5y = 0 (1)

and the 1MIN said:
PZ = k YZ (cause it needs to be colinear to it?) but then i'm introducing another variable and another equation is needed if i have to work it out simultaneously..
Title: Re: Maths Quest!!!!
Post by: Conic on April 26, 2014, 02:53:55 pm: Looking at this: Re: it's so hard :( it doesn't look like XP and XZ are perpendicular, so their dot product isn't 0. Also, you only have one variable, since you know y=7.
Title: Re: Maths Quest!!!!
Post by: bts on April 26, 2014, 02:58:40 pm: oh that one is a different question to the one i'm trying to solve. That one is like 3 something and the one i'm looking at is 6g) i believe. They are different questions
They are both similar and i've tried to apply what they did to this question but because before P lies on a straight line so you knew one of the coordinates already being 7, but with this one the P is on a slanting line so i'm not sure about any of the co ordinates making it hard.

the question was posted here: sigh... vectors!!!
Title: Re: Maths Quest!!!!
Post by: Conic on April 26, 2014, 04:00:27 pm: Yeah my bad, I was looking at the wrong question.

For that question you can use linear graphs.

The gradient of XZ is (5-0)/(3-6)=-5/3, to the gradient of a line perpendicular to XZ is -1/(-5/3)=3/5. The line XP has the gradient 3/5 (because it is perpendicular to XZ), and it passes through (3,5), so it's equation is y=3/5x+16/5.

The gradient of ZY is (5-0)/(9-6)=5/3, so the gradient of ZP is 5/3 (because P is on the line ZY). This line passes through (6,0), so it has the equation y=5/3x-10.

P is on the intersection of these lines, so to find the x coordinate we have 3/5x+16/5=5/3x-10, and find this x and substitute it into one of the equations to find y.

This gives x=99/8=12+3/8, and y=85/8=10+5/8.
Title: Re: Maths Quest!!!!
Post by: sujui on April 26, 2014, 04:10:05 pm: dear conic,

you're helping alot on this thread, I was wondering can you help prove how the medians of a triangle are concurrent? I know concurrent means that they intersect at one point only...
Title: Re: Maths Quest!!!!
Post by: Conic on April 26, 2014, 05:07:28 pm: I think this proves it:

(http://i.imgur.com/eBuaUA7.png)

For some triangle OAB, let the M be the midpoint of OA, let N be the midpoint of OB, and let X be the midpoint of AB. Let $\overrightarrow{OA}=\underset{\sim}{a}$ and $\overrightarrow{OB}=\underset{\sim}{b}$ . Let P be the point where the medians AN and BM intersect. We will show that O, P and X are collinear, which means that P is a point on the median OX that is also the point of intersection of the other two medians, which means the three medians meet at P i.e., they are concurrent.

AP is parallel to AN, so $\overrightarrow{AP}=\lambda \overrightarrow{AN}$ for some real number $\lambda$ . $\overrightarrow{AN}=\frac{1}{2}\underset{\sim}{a}-\underset{\sim}{b}$ , so we have $\overrightarrow{AP}=\lambda (\frac{1}{2}\underset{\sim}{b}-\underset{\sim}{a})$ .

BP is parallel to BM, so $\overrightarrow{BP}=\mu \overrightarrow{BM}$ for some real number $\mu$ . $\overrightarrow{BM}=\frac{1}{2}\underset{\sim}{a}-\underset{\sim}{b}$ , so we have $\overrightarrow{BP}=\mu(\frac{1}{2}\underset{\sim}{a}-\underset{\sim}{b})$ .

$\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}=\underset{\sim}{a}+\lambda(\frac{1}{2}\underset{\sim}{b}-\underset{\sim}{a})=(1-\lambda)\underset{\sim}{a}+\frac{\lambda}{2}\underset{\sim}{b}$

$\overrightarrow{OP}=\overrightarrow{OB}+\overrightarrow{BP}=\underset{\sim}{b}+\mu(\frac{1}{2}\underset{\sim}{a}-\underset{\sim}{b})=\frac{\mu}{2}\underset{\sim}{a}+(1-\mu)\underset{\sim}{b}$

$\implies (1-\lambda)\underset{\sim}{a}+\frac{\lambda}{2}\underset{\sim}{b}= \frac{\mu}{2}\underset{\sim}{a}+(1-\mu)\underset{\sim}{b}$

Since $\underset{\sim}{a}$ and $\underset{\sim}{b}$ are linearly independent we have $1-\lambda = \frac{\mu}{2}$ and $1-\mu=\frac{\lambda}{2}$ , which gives $\mu=\lambda=\frac{2}{3}$ .

Now we know that $\overrightarrow{OP}=(1-\frac{2}{3})\underset{\sim}{a}+\frac{\frac{2}{3}}{2}\underset{\sim}{b}=\frac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b})$ .

$\overrightarrow{OX}=\overrightarrow{OA}+\frac{1}{2} \overrightarrow{AB}=\underset{\sim}{a}+\frac{1}{2}(\underset{\sim}{b}-\underset{\sim}{a})=\frac{1}{2}(\underset{\sim}{a}+\underset{\sim}{b})$ .

Therefore O, P and X are collinear, as OP and OX are parallel and they share the point O. This means that P lies on the median OX. Since P is the point of intersection of the other two medians and it is on the median OX, all three medians meet at P, i.e., they are concurrent.