1. ## a vector question

Show that in 3-space the distance d from a point P to the line L through point A and B can be expressed as

d = (ll AP * AB ll)/ llABll

note: AP , AB are vectors.

Thank you very much.

2. Originally Posted by Jenny20
Show that in 3-space the distance d from a point P to the line L through point A and B can be expressed as

d = (ll AP * AB ll)/ llABll

note: AP , AB are vectors.

Thank you very much.
Are you asking for the formula for the distance from a point to a line in 3-dimensions.

3. Hi perfecthacker,

No. I think the question ask me to prove that d = (ll AP * AB ll)/ llABll.

4. Originally Posted by Jenny20
the question ask me to prove that d = (ll AP * AB ll)/ llABll.
Actually that formula is incorrect. It should be:
$d = \frac{{\left\| {\overrightarrow {AP} \times \overrightarrow {AB} } \right\|}}{{\left\| {\overrightarrow {AB} } \right\|}}.$
The formula contains the cross product. Cross products are associated with the sine of the angle between two vectors. Draw a diagram. See the sine of the angle between ${\overrightarrow {AP} }$ and ${\overrightarrow {AB} }$ is $\frac{{\left\| {\overrightarrow {AP} \times \overrightarrow {AB} } \right\|}}{{\left\| {\overrightarrow {AP} } \right\|\left\| {\overrightarrow {AB} } \right\|}}.$
Now if you solve for d, you will see how it works.

5. [QUOTE=Plato;33229]Actually that formula is incorrect. It should be:
$d = \frac{{\left\| {\overrightarrow {AP} \times \overrightarrow {AB} } \right\|}}{{\left\| {\overrightarrow {AB} } \right\|}}.$

Hi Plato,
Thank you very much.

This formula is correct and so your concept.
Could you please show me this question with picture? Thank you very much.
I know P, A , B should be formed a triangle. But I am not sure where exactly should be their correct position. Also does PB mean the d in the picture?

6. I can show you without a picture.

My technique uses something called "moment" (from physics).
Given a point $P(x_0,y_0,z_0)$ and a line $L$ (not on the line, otherwise there is nothing do to).

Consider the vector $\bold{u}=a\bold{i}+b\bold{j}+c\bold{k}$ to get aligned with the line $L$. And we can thing of vector $\bold{u}$ as some force.

Then the moment (without sign) created by the force $\bold{u}$ relative to point $P(x_0,y_0,z_0)$ is,
$F\cdot d$. Where $F$ is the magnitute, in this case $F=||\bold{u}||=\sqrt{a^2+b^2+c^2}$. And $d$ is the distance, what we are trying to find.

But there is another way to compute moment (without sign),
$|| \bold{r}\times \bold{u}||$
Where $\bold{r}$ is any vector drawn from the point to the line.
Thus,
$d\cdot ||\bold{u}||=||\bold{r}\times \bold{u}||$
Thus,
$\boxed{ d=\frac{||\bold{r}\times \bold{u}||}{||\bold{u}||} }$

This is my techiqnue (mnemonic), it makes more sense if you know what a moment is. I think it is easier like this.

7. Hi all,

I am taking Calculus course. Thus my solution is using calculus approach to solve it.

Thank you very much.

However, you have made some very specific assumptions.
First, you have assumed that B is the point such that $\overrightarrow {PB} \bot l$. That may not be the case.

Here is a way out. Assume that B’ is the foot of a perpendicular from P to l. Assume that A’ is another point on l. Now do exactly what you have done above using A' & B'. Thus you have d in terms of A’ & B’.
But $\overrightarrow {A'B'} = t\overrightarrow {AB}$, and so $\left\| {\overrightarrow {A'B'} } \right\| = \left| t \right|\left\| {\overrightarrow {AB} } \right\|$.

Thus you can make substitutions and have d in terms of A & B.

9. Originally Posted by Jenny20
Show that in 3-space the distance d from a point P to the line L through point A and B can be expressed as
d = (ll AP * AB ll)/ llABll
note: AP , AB are vectors.
Thank you very much.
Hello Jenny,

if you have two non-collinear vectors $\overrightarrow{AB}$ and $\overrightarrow{AP}$ then the result of the cross-product of these vectors is the normal vector $\overrightarrow{n}$ which is perpendicular to the two vectors.
And
the length of this normal vector is as great as area of the parallelogramm which is determined by the two vectors. (see attachment)

That means: $A_{parallelogram}=|\overrightarrow{AB} \times \overrightarrow{AP}|$
The area of the parallelogramm can also be calculated by:
$A_{parallelogram}=|\overrightarrow{AB}| \cdot d$ because d is the height in the parallelogram APCB.

So you get:
$|\overrightarrow{AB} \times \overrightarrow{AP}|=|\overrightarrow{AB}| \cdot d$. Now divide by $|\overrightarrow{AB}|$ and you'll get the formula which is used in the previous posts:

$d=\frac{|\overrightarrow{AB} \times \overrightarrow{AP}|}{|\overrightarrow{AB}|}$

EB

10. Hi earboth,

Thank you very much.

May I ask what program you use to make the drawing? And where can I download a copy of it? The outcome is excellent!

11. Originally Posted by Jenny20
Hi earboth,

Thank you very much.

May I ask what program you use to make the drawing? And where can I download a copy of it? The outcome is excellent!
To quote a mathematician,
It is the man and not the method.
.

Meaning, it is not the tools he has at his disposal (Calculus, Number theory,...). It is him, it is he who is good not what he has.

Perhaps, the same applies to earboth it is the man (his skill of drawing) but not the method (the program).

12. Originally Posted by Jenny20
Hi earboth,

Thank you very much.

May I ask what program you use to make the drawing? And where can I download a copy of it? The outcome is excellent!
Hello Jenny,

I use a program which is called "dynamic geometry". You'll find it at:

EB