Hi, having a little trouble on this vector question, any help appreciated!
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 $\displaystyle \color{blue}P = l(1)$
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.
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 $
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) $