Author Topic: most awesome question thread :) (Read 8299 times) Tweet Share

TrueTears · « **Reply #75 on:** January 14, 2010, 04:02:30 pm »

Quote from: kamil9876 on January 14, 2010, 01:32:02 pm

Just to illustrate how vectors can trivialise geometry:

let $a$ and $b$ be two adjacent sides of the parralogram.

We have that

$|a+b|=|a-b|$

$|a+b|^2=|a-b|^2$

$(a+b).(a+b)=(a-b).(a-b)$

$a.a+2a.b+b.b=a.a-2a.b+b.b$

$4a.b=0$

$a.b=0$

Thus $a$ and $b$ are perpendicular as required.

lol, nice, I remember you showing me this longggg time ago

mangopop · « **Reply #76 on:** January 14, 2010, 09:39:05 pm »

Thank you brightsky and kamil!

kamil9876 · « **Reply #77 on:** January 14, 2010, 09:59:36 pm »

Quote

lol, nice, I remember you showing me this longggg time ago

lol serious, i can't remember ever doing this. I had to think ~~a bit~~ this time around.

superflya · « **Reply #78 on:** January 15, 2010, 12:21:10 am »

^^ being a tad modest

kamil9876 · « **Reply #79 on:** January 15, 2010, 12:31:00 am »

Lol I didn't mean a bit literally

superflya · « **Reply #80 on:** January 15, 2010, 12:36:57 am »

superflya · « **Reply #81 on:** January 17, 2010, 06:20:23 pm »

find area enclosed by $y= \sqrt{x}$ & $y=6-x$ & $y=1$

im not sure as to the order of equations that i have to subtract, its usually top part of area minus bottom but now theres right left and bottom?

Edit: got it -.-

superflya · « **Reply #82 on:** February 17, 2010, 10:38:46 pm »

find the point P on the line $x-6y=11$ such that $\underset{OP}{\rightarrow}$ is parallel to the vector $3i+j$

QuantumJG · « **Reply #83 on:** February 17, 2010, 11:29:03 pm »

What I did was:

Let P = (x,y) where x,y are a point on the line.

$\vec{OP}$ = (x,y) - (0,0) = (x,y)

$\vec{OP}$ . (3,1) = 1

$\therefore (x,y) . (3,1) = 1 \rightarrow 3x + y = 1$

Now $x = 11 + 6y \rightarrow 3(11 + 6y) + y = 1$

$\therefore 19y + 33 = 1 \rightarrow y = - \dfrac{32}{19}$

$\therefore x = \dfrac{17}{19}$

$\therefore P = ( \dfrac{17}{19}, \dfrac{-32}{19} )$

superflya · « **Reply #84 on:** February 17, 2010, 11:35:58 pm »

book says $(-11, \frac{-11}{3})$ :S

GerrySly · « **Reply #85 on:** February 18, 2010, 12:08:20 am »

Assuming $a$ is the x-ordinate of our point...

$P(a, \frac{a-11}{6})$ , now we get the gradient of that point and make it equal $\frac{1}{3}$ because that is the gradient of the vector (worded wrong I assume, but that's what I was thinking heh)

$\begin{align*} \frac{\frac{a-11}{6}}{a}&=\frac{a-11}{6a}\\ \frac{1}{3}&=\frac{a-11}{6a}\\ 2a&=a-11\\ a&=-11 \end{align*}$

Sub that back in

$P(-11, -\frac{11}{3})$

QuantumJG · « **Reply #86 on:** February 18, 2010, 12:36:15 am »

Quote from: superflya on February 17, 2010, 11:35:58 pm

book says $(-11, \frac{-11}{3})$ :S

I found a much simpler way to solve it and don't listen to my method above.

The solution is easy.

First let's draw a line through the point (3,1) which is $y = \dfrac{x}{3}$ and let's find where $y = \dfrac{x}{6} - \dfrac{11}{6}$ intersects $y = \dfrac{x}{3}$ . Why? Well because the point could always make the vector anti-parallel to (3,1).

So

$\dfrac{x}{6} - \dfrac{11}{6} = \dfrac{x}{3} \rightarrow -\dfrac{x}{6} = \dfrac{11}{6} \rightarrow x = -11$ & $y = \dfrac{11}{3}$

So P = (-11, $\dfrac{11}{3}$ )

superflya · « **Reply #87 on:** February 18, 2010, 07:48:12 pm »

-.- all the working and the solution was that straightforward \facepalms

legend quantum

superflya · « **Reply #88 on:** February 25, 2010, 06:04:38 pm »

the number of real solutions for $x^{4}-x^{3}=\csc ^{2}(x)-\cot ^{2}(x)$ is?

moekamo · « **Reply #89 on:** February 25, 2010, 07:28:09 pm »

Quote from: superflya on February 25, 2010, 06:04:38 pm

the number of real solutions for $x^{4}-x^{3}=\csc ^{2}(x)-\cot ^{2}(x)$ is?

$x^{4}-x^{3}=\csc ^{2}(x)-\cot ^{2}(x)$
$\implies x^{4}-x^{3}=1$ (since $1+\cot^2x=\csc^2x \implies 1=\csc^2x - \cot^2x$ )

graphs the curves $y=x^{4}-x^{3}, y=1$ on calc and see how many intersections there are

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