The question goes like this:
A point $A$ has position vector $2\mathbf{i}+3\mathbf{j}$ and the line $L$ has equation: $$\vec{r}=5\mathbf{i}+6\mathbf{j}+3\mathbf{k}+t(2\mathbf{i}+2\mathbf{j}-\mathbf{k})$$
1) Find the position vector of the point $P$ on $L$ such that $\vec{AP}$ is perpendicular to $L$.
The question is pretty long but if i can find $P$ I can do the rest:
I concluded that the point $P$ should have position vector $x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ and that $\vec{AP}$ should be:
$$=(x\mathbf{i}+y\mathbf{j}+z\mathbf{k})-(2\mathbf{i}+3\mathbf{j})$$
If $\vec{AP}$ is perpendicular to $L$, then the direction vector $(2\mathbf{i}+2\mathbf{j}-\mathbf{k})$ of $L$ should be perpendicular to $\vec{AP}$:
$$\cos 90°=\frac{(2\mathbf{i}+2\mathbf{j}-\mathbf{k})\cdot [(x\mathbf{i}+y\mathbf{j}+z\mathbf{k})-(2\mathbf{i}+3\mathbf{j})]}{|2\mathbf{i}+2\mathbf{j}-2\mathbf{k}||(x\mathbf{i}+y\mathbf{j}+z\mathbf{k})-2\mathbf{i}+3\mathbf{j}|}$$
$$2\mathbf{i}+2\mathbf{j}-\mathbf{k}\cdot [(x\mathbf{i}+y\mathbf{j}+z\mathbf{k})-(2\mathbf{i}+3\mathbf{j})]=0$$
But solving for $x$, $y$ and $z$,
$P$ has the same value as $A$ i.e $2\mathbf{i}+3\mathbf{j}$
Help please
Lee.
Best Answer
Let $A(2,3)$ have position vector $\vec{a} = <2,3,0>$
Let the line be $r = <5,6,3>+\ t<2,2,-1>$
Then you have that:
$$(\vec r - \vec a)\cdot \vec r =0$$
or rather you can also say:
$$(\vec r - \vec a)\cdot<2,2,-1>=0$$
From here, you solve and find the value of $t$, and hence the position vector of the needed point.
$$6+4t+6+4t-3+t=0$$
$$\therefore t=-1$$
So the position vector of the point is $<3,4,4>$