1. ## Vectors question!

Hi, having a little trouble on this vector question, any help appreciated!

2. Write the equation of the line determined by $\displaystyle A\;\&\;B$: $\displaystyle l(t) = B + t\left( {\overrightarrow {AB} } \right)$.
Be sure to note that $\displaystyle \overrightarrow {AB} = B - A$. The point you want is $\displaystyle \color{blue}P = l(1)$

3. Why is the point I want where t=1? I thought as AP=2BP I would want to set t=(2/3)?

4. My text says that P= (-7i, -2j, -5k)
Is this correct?

5. If you use this equation $\displaystyle l(t) = B + t\left( {\overrightarrow {AB} } \right)$, then $\displaystyle l(1) = \left\langle { - 7, - 2, - 5} \right\rangle$, as in your textbook.
That means that the order of the points is $\displaystyle A-B-P$, B is between A & P.

6. Ok, thank you, I think I see now, they could be in that order from the way the question its written. Not sure why I assumed P was between them.

7. you are asked to find position vector $\displaystyle \overrightarrow{OP}$

$\displaystyle \overrightarrow{OP} = \overrightarrow{OB} + \overrightarrow{BP} = \overrightarrow{OA} + \overrightarrow{AP}$

$\displaystyle \overrightarrow{OA} = 5i + 4j + k$

$\displaystyle \overrightarrow{OB} = -i + j - 2k$

$\displaystyle \overrightarrow{AB} = -\overrightarrow{OA} + \overrightarrow{OB} \equiv (-i + j - 2k) - (5i + 4j + k) \equiv -6i - 3(jk)$

$\displaystyle AB = 2BP + PB \equiv 3PB$

$\displaystyle \overrightarrow{AP} = \frac{2}{3} (-6i -3(jk)) \equiv -4i - 2(jk)$

$\displaystyle \overrightarrow{OP} = \overrightarrow{0A} + \overrightarrow{AP} \equiv (5i + 4j + k) + (-4i - 2(jk)) \equiv i + 2j - k$

i think this is load of rubbish but i try

corrected for LHS and Plato's comments

$\displaystyle \overrightarrow{OP} = \overrightarrow{OB} + \overrightarrow{BP} \text{ where } \overrightarrow{BP} = \overrightarrow{AB}$

from above

$\displaystyle (-i + j - 2k) + (-6i - 3j - 3k) = -7i -2j -5k$

8. Originally Posted by sammy28
you are asked to find position vector OP
$\displaystyle \color{red}OP = OA + PB = OA + AP$
You assume alot in the above.

9. Sammy I think you've done what I've done and assumed that P is in between A and B, I think P is one more lot of AB from B.

10. Originally Posted by LHS
Hi, having a little trouble on this vector question, any help appreciated!
There are 2 points satisfying the given conditions:

$\displaystyle P_1 \ at\ t = \frac13$

$\displaystyle P_2 \ at\ t =- \frac12$

11. Hello, LHS!

Here's an "eyeball" solution . . .

The position vectors of points $\displaystyle A$ and $\displaystyle B$ relative to an origin $\displaystyle O$
. . are: .$\displaystyle \vec A \:=\: 5i + 4j + k,\;\;\vec B \:=\: \text{-}i + j - 2k$.

Find the position vector of the point $\displaystyle P$ which lines on $\displaystyle AB$ produced
. . such that: .$\displaystyle \overrightarrow{AP} \:=\:2\cdot\overrightarrow{BP}$.

. . $\displaystyle \begin{array}{ccccc} (5,4,1) & & (\text{-}1,1,\text{-}2) & & (?,\,?,\,?) \\ * & \longrightarrow & *& \longrightarrow & * \\ A & & B & & P \end{array}$

Going from $\displaystyle A$ to $\displaystyle B$, we see:

. . $\displaystyle \begin{array}{ccc} \text{variable} & \text{moves} & \text{change} \\ \hline x & \text{5 to -1} & \text{-}6 \\ y & \text{4 to 1} & \text{-}3 \\ z & \text{1 to -2} & \text{-}3 \end{array}$

Going from $\displaystyle B$ to $\displaystyle P$, the change would be the same:

. . $\displaystyle \begin{array}{ccc} B & \text{change} & P \\ \hline x=\text{-}1 & \text{-}6 & x = \text{-}7 \\ y = 1 & \text{-}3 & y = \text{-}2 \\ z = \text{-}2 & \text{-}3 & z = \text{-}5 \end{array}$

Threfore: .$\displaystyle P \:=\:(\text{-}7,\text{-}2,\text{-}5)$

12. Ah, thank you, that is much clearer!

13. Originally Posted by LHS
Ah, thank you, that is much clearer!
Your textbook gives the answer that Soraban & I both gave you.
Reading the question as a mathematician, from the wording I assumed that if order made a difference. That is the point P could not be between A & B. B would be between A & P.