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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.

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  • $\begingroup$ You last equation describes a plane that passes through $2\mathbf i+3\mathbf j$ and is perpendicular to the line. You now have to find the intersection of the two. $\endgroup$ – amd Apr 4 '17 at 18:07
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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 $<3,4,4>$

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  • $\begingroup$ Thanks a lot samjoe $\endgroup$ – Lee Apr 4 '17 at 16:56
  • $\begingroup$ I was able to solve for t=-1 $\endgroup$ – Lee Apr 4 '17 at 16:56
  • $\begingroup$ I suppose the approach you are using doesn't take position vector to be on the line, so it will naturally lead to two solutions, one such that the magnitude of vector difference with (2,3) be zero, which leads to same point (2,3). $\endgroup$ – samjoe Apr 4 '17 at 17:02
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$(5,6,3)$ 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

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

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  • $\begingroup$ @samjoe thanks. $\endgroup$ – Doug M Apr 4 '17 at 16:51

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