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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: airline on March 14, 2009, 02:30:22 pm

Title: Vector Revision Help
Post by: airline on March 14, 2009, 02:30:22 pm: We were given a vector revision sheet.
I think it is out of 'smartstudy' guide.

Can someone please show me how to do these...was going really well at the start but now I have hit a wall.
Some I know but can't prove them.

1. The position vector of point P is given by $OP=2cos(\Theta)\tilde{i} + (1-2cos(\Theta))\tilde{j}$ . OP is perpendicular to the line $x+y=1$ if

A: $cos(\Theta)=\frac{1}{4}$
B: $cos(\Theta)=\frac{1}{2}$
C: $cos(\Theta)=\frac{1}{\sqrt{2}}$
D: $cos(\Theta)=\frac{\pi}{4}$
E: $cos(\Theta)=\frac{-1}{4}$

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2. The position vector $\tilde{r}(t)$ of a particle at time $t\;(t\geqslant0)$ is given by $\tilde{r}(t)=(2t-1)\tilde{i} + t^2\tilde{j}$ . The Cartesian equation of the curve along which the particle moves is:

A: $y=\frac{1}{2}(x+1)^2, x\geqslant-1$
B: $y=\frac{1}{2}(x+1)^2, x\geqslant0$
C: $y=(\frac{x+1}{2})^2, x\geqslant1$
D: $y=(\frac{x+1}{2})^2, x\geqslant0$
C: $y=(\frac{x+1}{2})^2, x\geqslant-1$

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3. The position vector, $\tilde{r}(t)$ , of a particle at any time $t\;(t\geqslant0)$ is given by $\tilde{r}(t)=(h+acos(t))\tilde{i} + (k+bsin(t))\tilde{j}, \quad a,b,h,k \in R$ and $a \neq b$ . The curve along which this particle moves is:

A: a straight line
B: a parabola
C: an ellipse
D: a circle
E: an hyperbola

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4. OABC is a rhombus in which $OA=CB=\tilde{a}$ and $AB=OC=\tilde{b}$ .

Which one of the following could be used to prove that the diagonals of a rhombus are at right angles to each other?

A: $\tilde{a}.\tilde{b}=0$
B: $(\tilde{a}+\tilde{b}).(\tilde{a}+\tilde{b})=0$
C: $(\tilde{a}+\tilde{b}).(\tilde{a}-\tilde{b})=0$
D: $(\tilde{a}+\tilde{b}).(\tilde{a}+\tilde{b})=\frac{\pi}{2}$
E: $(\tilde{a}+\tilde{b}).(\tilde{a}-\tilde{b})=\frac{\pi}{2}$

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5. A particle moves in the x-y plane so that its position vector $\tilde{r}(t)$ is given by $\tilde{r}(t)=acos(nt)\tilde{i} + asin(nt)\tilde{j}$ .
Which one of the following is incorrect

A: the particle travels in a circular path.
B: the speed of the particle is not constant.
C: the particle starts its motion at (a,0).
D: the period of the motion is $\frac{2\pi}{n}$ .
E: the acceleration is always directed towards the origin.
Title: Re: Vector Revision Help
Post by: /0 on March 14, 2009, 06:14:38 pm: 1. If a vector is perpendicular to the line $y=1-x$ , it will also be perpendicular to the line $y=-x$ . The position vector of any point on the line $y=-x$ is $\mathbf{r}=m(\mathbf{i}-\mathbf{j})$ .
So $\vec{OP} \cdot \mathbf{r}=(m)(2\cos{\theta})+(-m)(1-2\cos{\theta})=0$

$\Rightarrow 2\cos{\theta}-(1-2\cos{\theta})=0$

$4\cos{\theta}=1$

$\cos{\theta}=\frac{1}{4}$

Hence, A

2.

$x=2t-1$ (1)

$y=t^2$ (2)

Solving for t in (1),

$t = \frac{x+1}{2}$

Subbing it into (2):

$y=\left(\frac{x+1}{2}\right)^2$

The range of $x=2t-1$ gives the domain of $y=\left(\frac{x+1}{2}\right)^2$

Since $t \geq 0$ , the range of x is $[-1,\infty)$ , so E is the correct answer.

3.

$x=h+a\cos{t}\Rightarrow \cos{t}=\frac{x-h}{a}$

$y=k+b\sin{t}\Rightarrow \sin{t} = \frac{y-k}{b}$

Using the identity $\sin^2{t}+\cos^2{t}=1$ ,

$\left(\frac{x-h}{a}\right)^2+\left(\frac{y-k}{b}\right)^2=1$

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

It is an ellipse, since $a\neq b$ .

Hence C.

4.

For diagonals to be perpendicular, we need

$\vec{OB}\cdot \vec{CA}=0$

$\vec{OB}=\mathbf{a}+\mathbf{b}$

$\vec{CA} = \mathbf{a}-\mathbf{b}$

$\therefore (\mathbf{a}+\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})=0$

Hence, C.

5.

$x=a\cos{(nt)}\Rightarrow \cos{(nt)} = \frac{x}{a}$

$y=a\sin{(nt)}\Rightarrow \sin{(nt)} = \frac{y}{a}$

Using $\sin^2{(nt)}+\cos^2{(nt)}=1$ ,

$\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$

$x^2+y^2=a^2$

So A is valid,

$v(t) = -na\sin{(nt)}\mathbf{i}+na\cos{(nt)}\mathbf{j}$

$|v(t)| = (-na\sin{(nt)})^2+(na\cos{(nt)})^2 = n^2a^2$ , i.e. speed is constant.

So B is valid,
D is also valid, since the period of both functions is $\frac{2\pi}{n}$ .

$a(t) = -n^2a\cos{(nt)}\mathbf{i}-n^2a\sin{(nt)}\mathbf{j}$

But $x(t) = a\cos{(nt)}, y(t) = a\sin{(nt)}$ ,

$\Rightarrow a(t)=-n^2x(t)\mathbf{i}-n^2y(t)\mathbf{j}=-n^2(r(t))$

So the acceleration vector is always opposite the position vector, i.e. pointing towards the origin.

With C, you don't know the domain of t, so you can't assume it will begin at (a,0).

Hence, C.
Title: Re: Vector Revision Help
Post by: airline on March 15, 2009, 12:30:49 pm: Thankyou /0! You are amazing!

One more I need to ask for help....

A long distance swimmer leaves the starting point O of a race and swims 2km due south to marker A, 4km on a bearing N $45^\circ$ E to marker B and then 3km due east to marker C.

Given $\tilde{i}$ and $\tilde{j}$ are unit vectors of magnitude 1km in the east and north directions respectively, the long distance swimmer's position vector at marker C relative to O is:

A: $5.57\tilde{i}+1.06\tilde{j}$
B: $5.57\tilde{i}+5.06\tilde{j}$
C: $0.57\tilde{i}+6.06\tilde{j}$
D: $6.06\tilde{i}+0.57\tilde{j}$
E: $1.06\tilde{i}+5.57\tilde{j}$
Title: Re: Vector Revision Help
Post by: kamil9876 on March 15, 2009, 02:57:15 pm: Refer to attatched diagram.
So at A he is at the point (0,-2) since he swam 2km in the -i direction. Travelling from A to B, his i component changes a bit and so does his j component. U hav to get the vector into i and j components as i did below using trig and then add it into his i and j components. The final vector is easy, u just hav to add on 3km to his i component of his position vector at B.