pretty vectorial :D

[kingjewel1]

Hi guys

I'm on my vectors again.

Last time you really helped me understand what i was doing!

 

Refered to a fixed origin A=(5i-j-k) B=(i-5j+7k)

Find the position vector of the point D, where Dis not A on AB such that

|OD|=|OA|

 

i did

OD= xi+yj+zk

from r=a+t(B-A)

x=5-4t

y=-1-4t

z=-1+8t

 

|AD|^2=(x-5)^2+(y+1)^2+(z+1)^2

|AD|^2=(5-4t-5)^2+(-1-4t+1)^2+(-1+8t+1)^2

d(|AD|^2)/dt=(32+32+128)t

am i using the right method?

i also did Cos(OAOD) but got silly decimals. for the vector eq.

 

Thanks in advance!

[timo]

Starting to write down the equation for the line that D lies on (your r = A + t(B-A)) is a good start. But why you then proceed with calculating anything on |AD| is completely unclear to me.

Unless you made a typo somewhere, it´s |OD| = |OA| that you have as 2nd condition to fix the point D, not a condition involving |AD|. The straightforward solution is to start from |OD| = |OA|. |OA| is known because you know the coordinates of A, |OD| can be expressed by the line r(t) you allready got. Solve for t. You´ll get two possible results for t. One will be such that r(t)=A which was excluded in the question. The other one is the one you´re looking for, then.

As a hint here: As distances are allways positive, the condition |OA|² = |OD|² is the same as |OA| = |OD| and eleminates the annoying sqare root.

 

Differentiation (what you did) is often used to find extremal values (the differential is zero, then). So if you use d(|AD|²)/dt = 0 and solve it, you find the point D that closest to A. Not much of a surprise that you´d get t=0 and the point closest to A would be A itself. But that´s not what is asked for.

 

There´s an alternative way to solve the question which is quite elegant in my opinion but needs a bit of understanding of vector math and might be a bit more work to actually do. But since it might help you understand the problem and you might also find it interesting, I´ll sketch it here:

Given an abritrary vector k, any vector v can be split up into a part parallel to k and a part perpendicular to k in a unique way. In your case, you could split up the vector a which is the coordinates of A into a part a_par parallel to the line through AB and a part a_perp which is perpendicular to that line. a_perp will hit the line at a point P. This is the point on the line which is closest to the origin. If you add a_par to that point, you´ll end up on point A, of course. But if you substract a_par from there, you´ll end up on another point which has the same distance from the origin as A (because |a_perp|² + |a_par|² = |a_perp|² + |-a_par|² <-- phytagoras). As a_par is parallel to the line so is -a_par. The point you got by D = a_perp - a_par is the one you´re looking for.

This method is probably best understood if you draw it on a piece of paper, so you might do that after you solved the question the straightforward way proposed above using the correct coordinates, then.

[kingjewel1]

Thank you very much Atheist. It helped me greatly.

so i did

[math]|OA|=sqrt(5^2+1^2+1^2)[/math]

since |OD|=|OA|

from above x,y,z

[math]|OD|^2=(x-5)^2+(y+1)^2+(z+1)^2[/math]

[math]27=16t^2+16t^2+64t^2[/math]

[math]t=sqrt(9/32)[/math]

 

[math]5-4t=2.87[/math]

[math]-1-4t=-3.12[/math]

[math]-1+8t=3.24[/math]

so therefore [math]d=3i-3j+3k[/math]

i must have made a mistake because i didn't think they were supposed to be decimals :S

 

Your second method is interesting. i'll post my workings.

Cheers

[timo]

If you have a point P=(x, y, z) then it´s squared distance from the origin is x²+y²+z². Whether these x,y,z depend on another parameter t or doesn´t matter. So why did you use "|OD|² = (x-5)² + (y+1)² + (z+1)²" instead of "|OD|² = x²+y²+z²" ?

The rest of your math seems ok, so try it again with |OD|² = x²+y²+z².

[kingjewel1]

---

[kingjewel1]

Yes my mistake, i took the equation from |AD|

i should have done

[math]x=5-4t[/math]

[math]y=-1-4t[/math]

[math]z=-1-8t[/math]

this gives me t=1/2 ^t=0

thanks

 

For the next method,

I used [math]A (5i-j-k) B(i-5j+7k)[/math]

// a_par to AB

[math]r=5i-j-k+t(i-5j+7k)[/math]

 

a_perp

A.(x,y,z)=0

B.(x,y,z)=0

if z=1

then

[math]r=1/2i+2/3j+k[/math]

 

would these be correct?

if sqrt27=a_perp-a_par how do i solve if r is in terms of t?

[timo]

Not much time to answer right now since I have to go to work (damn, I´m allready quite late). But I drew a sketch for you that you might like. I´ll check back here later this evening.

[kingjewel1]

Ah that looks good!

I did

AB.OP=0, P=x,y,z

(-4*x)+(-4*y)+(8*z)=0

-24+96t=0

t=1/4

if D is twice away as P then

t=1/2

therefore

3i-3j+3k

 

Thank you Atheist