Finding the position vector of a point $P$ on a line $L$ such that $\vec{AP}$ is perpendicular to $L$.

Question

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.

Answer 1

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 $<3,4,4>$

Answer 2

$(5,6,3)$ is a point on the line $L.$ Lets call this point $B.$

Find the vector $\mathbf v = \vec {BA}$

$\mathbf v = (2,3,0) - (5,6,3) = (-3,-3,-3)$

$\frac {\mathbf v\cdot (2,2,-1)}{\|\mathbf v\|\|(2,2,-1)\|}$ gives us the cosine of the angel between $\mathbf v$ and the direction vector of the line.

$\frac {\mathbf v\cdot (2,2,-1)}{\|(2,2,-1)\|}$ would be the distance you need to travel from $B$ to get to $P$

$P=\frac {\mathbf v\cdot (2,2,-1)}{\|(2,2,-1)\|^2} (2,2,-1) + (5,6,3)$

$P=$$\frac {(-3,-3,-3)\cdot (2,2,-1)}{2^2+2^2+(-1)^2)} (2,2,-1) + (5,6,3)\\ -(2,2,-1) + (5,6,3) = (3,4,4)$

Alternative

$2\mathbf{i}+2\mathbf{j}-\mathbf{k}\cdot [(x\mathbf{i}+y\mathbf{j}+z\mathbf{k})-(2\mathbf{i}+3\mathbf{j})]=0\\ 2(x-2) + 2(y-3) - z = 0$

and

$x = 5+2t\\ y = 6+2t\\ z= 3-t$

Substituting

$2(3+2t) + 2(3+2t) - (3-t) = 0\\ t = -1\\ (x,y,z) = (3,4,4)$