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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: gta007 on April 02, 2008, 05:09:23 pm

Title: MY Vector Problems
Post by: gta007 on April 02, 2008, 05:09:23 pm: My turn to post some vector problems which I have been unable to answer:

Q1) Find the point P on the line $x - 6y = 11$ such that $\vec{OP}$ is parallel to the vector 3i + j.

Q2) Show that, if a vector in three dimensions make angles $\alpha$ , $\gamma$ and $\beta$
respectively with the x, y and z axes then $cos^2\alpha + cos^2\gamma + cos^2\beta = 1$
Title: Re: MY Vector Problems
Post by: Mao on April 02, 2008, 06:08:30 pm: Q1)
$y=\frac{x-11}{6}$

then the point P has the coordinates $\left( x,\frac{x-11}{6} \right)$ , and the vector OP will be $xi+\frac{x-11}{6}j$

for it to be parallel to $3i+j$ , there must be a constant $k$ when multiplied to OP, makes $3i+j$

or in other words, the vector OP must have the same ratio of i to j

hence, $\frac{3}{1}=\frac{x}{\frac{x-11}{6}}$

and the rest is trivial

Q2

if you are using essentials book, theres a diagram on P71 that will help u:

for a vector $a=a_1i+a_2j+a_3k$

$\cos \alpha=\frac{a_1}{|a|}$ and etc...

that means:

$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma$

$\implies \frac{(a_1)^2}{|a|^2}+\frac{(a_2)^2}{|a|^2}+\frac{(a_3)^2}{|a|^2}$

$\implies \frac{(a_1)^2+(a_2)^2+(a_3)^2}{|a|^2}$

$\implies \frac{(a_1)^2+(a_2)^2+(a_3)^2}{(a_1)^2+(a_2)^2+(a_3)^2}$

$\therefore \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma=1$
Title: Re: MY Vector Problems
Post by: gta007 on April 02, 2008, 07:19:47 pm: Thank you very much :)
Title: Re: MY Vector Problems
Post by: gta007 on April 03, 2008, 09:58:14 am: Okay more questions, I seem to have problems with the same types of question.

eg. A and B are defined by the position vectors a = 2i - 2j - k and b = 3i + 4k
Find the unit vector which bisects <AOB.

I already have $a\hat = 1/3(2i -2j - k) and b\hat = 1/5(3i + 4k)$

I found c, half way between a and b to be 2.5i - j + 1.5k. I don't know what to do from there.

The next question is more of the same.....
eg, A and B are points defined by the position vectors a = i + 3j - k and b = j + k
I have found the vector resolute of a in direction of b as j + k.

Then it tells me to find the unit vector through A perpendicular to OB. I have no idea.

Thirdly and lastly, I have trouble answering the questions which say find the shortest distance from x to line y etc.....

eg. A, B and C are points defined by the position vectors a = i + 2j + k, b = 2i + j - k and c = 2i - 3j + k

I found $\vec{AB}$ as i - j - 2k and $\vec{AC}$ as i - 5j
I then went on to find the vector resolute of $\vec{AB}$ in the line $\vec{AC}$ as $3/13(i -5j)$
It then asks me to find the shortest distance from B to line AC. Here is where I am stuck.

Any help to help clarify these problems would be much appreciated.
Title: Re: MY Vector Problems
Post by: Mao on April 03, 2008, 02:12:02 pm: 1) <---omg cant believe i didnt realise it...
now that you have your two unit vectors, imagine them being the two sets of opposite sides of a rhombus. The angular bisector is easily the sum of the unit vectors, and then just find the unit vector of that :D

2)
vector resolutes are basically breaking vector a into rectangular components with b as an axis, which means:
a in the direction of b + a perpendicular to b = a

3)
this question follow the same logic as 2)
the shortest distance from B to AC is when the path is perpendicular to AC, or the vector of AB perpendicular to AC.
Title: Re: MY Vector Problems
Post by: Mao on April 03, 2008, 03:17:20 pm: solutions:

1)
$\hat{a}=\frac{2i-2j-k}{3}$ and $\hat{b}=\frac{3i+4k}{5}$

then the angular bisector can be found by arranging $\hat{a}$ and $\hat{b}$ like a rhombus, then finding the diagonals:

$c=\hat{a}+\hat{b}=\frac{19i-10j-27k}{15}$

$\therefore \hat{c}=\frac{3}{238}(19i-10j-27k)$

(why does that number look too complicated...?)

2)
let u be the vector resolute of $a$ parallel to $b$ :

$u=j+k$

let w be the vector of a perpendicular to $\vec{OB}=b$

$w+u=a$

$w=a-u=(i+3j-k)-(j+k)=i+2j-2k$

3)
let u be the vector resolute of $\vec{AB}$ in the direction of $\vec{AC}$

$u=\frac{3}{13}(i-5j)$

let w be the vector of $\vec{AB}$ perpendicular to $\vec{AC}$

$w+u=\vec{AB}$

$w=\vec{AB}-u=(i-j-2k)-\frac{3}{13}(i-5j)=\frac{10i-28j-26k}{13}$

shortest distance $=|w|=\frac{2\sqrt{390}}{13}\approx 3.04$ units

(why does this number also look too complicated...?)