Author Topic: Maths Quest!!!! (Read 2244 times) Tweet Share

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?

Conic · « **Reply #1 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 $\underset{\sim}{u}=\frac{1}{\sqrt{2}}\underset{\sim}{i}-\frac{1}{\sqrt{2}}\underset{\sim}{k}$

bts · « **Reply #2 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?

Conic · « **Reply #3 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.

bts · « **Reply #4 on:** April 25, 2014, 09:38:29 pm »

would you assume E is half OB cause otherwise not sure how to do it

Conic · « **Reply #5 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 $\frac{10}{3\sqrt{2}}$ .

bts · « **Reply #6 on:** April 25, 2014, 09:46:32 pm »

the ans says its 5/(3squareroot2) though?

Conic · « **Reply #7 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.

bts · « **Reply #8 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?

Conic · « **Reply #9 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.

bts · « **Reply #10 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 $\frac{10}{3\sqrt{2}}$ .

Conic · « **Reply #11 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 $\left |1\times -\frac{10}{3\sqrt{2}} \right |=\frac{10}{3\sqrt{2}}$ .

bts · « **Reply #12 on:** April 25, 2014, 10:22:26 pm »

thank you so so much :
just curious: how did they get the (5squareroot2) / 3

Conic · « **Reply #13 on:** April 25, 2014, 11:05:35 pm »

By rationalising the denominator:

$\frac{10}{3\sqrt{2}}=\frac{10}{3\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{10\sqrt{2}}{3\times 2}=\frac{5\sqrt{2}}{3}$

bts · « **Reply #14 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

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