vectors
The position vectors of points $A$ and $B$ relative to an origin $O$ are $5\hat i+4\hat j+\hat k, \, -\hat i+\hat j-2\hat k$ respectively. Find the position vector of the point $P$ which lies on $AB$ produced such that $AP=2BP$.
I'm struggling to understand the sketch given in the solution of my problem. The question states the point $P$ lies on the vector $AB$, which means $P$ should be between the points $A$ and $B$ right?
Let $\mathbf{OP} = <a,b,c> $.
We know that $AB$ and $\mathbf{OP}$ must be perpendicular, so the dot product of $\mathbf{OP}$ and $<2,2,-2>$ is $0$ ($<2,2,-2>$ plays the role of $\mathbf{v}$ in the diagram below)
$$<2,2,-2> \bullet <a,b,c> = 2a+2b-2c = 0$$ $$ \Rightarrow a+b-c=0$$ We also have that for some value of $\lambda$ (let's just call it $t$) $$<1,2,2> + <2,2,-2>t $$ $$=<1+2t, 2+2t, 2-2t> = <a,b,c>$$ because $P$ lies in the $AB $
Combining the two equations, we get $$1+2t +2 +2t - 2 +2t = 0 \Rightarrow 6t = -1$$ $$\Rightarrow t = -1/6$$
So $\mathbf{OP} = <2/3, 5/3, 7/3> = \frac13 <2,5,7>$
At each point in the space corresponds a position vector and viceversa:
$$P(x,y)\equiv OP=x \vec i+y\vec j$$
When you consider two points
$$P(a,b)\equiv OP=a \vec i+b\vec j \quad Q(c,d)\equiv O Q=c \vec i+d\vec j$$
the vector from $P$ to $Q$ is given by
$$PQ=P-Q=(a-c,b-d)=(a-c) \vec i+(b-d)\vec j$$
From a mathematical point of view vector $PQ$ is also a vector form the origin (in a vector space all vectors are form the origin)
$$PQ=OR=(a-c,b-d)=(a-c)\vec i+(b-d)\vec j$$
such that $$OP+OR=OQ$$
Anyway you could also though to $PQ$ as a vector applied in $P$ such that:
$$OP+PQ=OQ$$
and in this sense they can be moved like free vectors.
Best Answer
Hint: let $\overrightarrow{BP} = q$ then $\overrightarrow{AP} = 2q$ by construction.
Then $p=a+2q=b+q\,$, and eliminating $q$ between the two equations gives $p$ in terms of $a,b\,$.