The smallest distance between two vectors?

A radar station tracks a jet fighter flying with constant speed. If the radar station is considered to be at the origin, the fighters starting position is 2i + 8j + k and 1 minute later it is at 8i - 4j + 13k. (therefore, velocity = 6i - 12j + 12k)

Find the point along the path where the fighter is closes to the station?

So far, I've found the velocity of the fighter and therefore the vector that describes its position at any time (using m), i.e.

(2 + 6m)i + (8 - 12m)j + (1 + 12m)k

I then used my CAS to find m when the dot product of
(2 + 6m)i + (8 - 12m)j + (1 + 12m)k . (0i + 0j + 0k) = 0 and solve for m, but I got no answer (only 'true').

I'm now lost for as how to solve this question, please help!

Hey iamapineapple.

Hint: The smallest distance between two straight lines is at the points where the lines are perpendicular to each other.

Thankyou, but I did say I tried that with the dot product.

You should find the normal vector and then use that to get the answer.

Hey Chiro, I'm still kind of lost? So I should find the vector resolute of on vector parallel to another? If I did that with 0, wouldn't I get infinite answers? :/
Thanks for the help so far though

Basically I'm advocating you look at the skew-line solution:

Math Forum - Ask Dr. Math

I still don't know what to do. Could you propose a solution please?

Iama...

Please follow me carefully.

the method you used is good for any point other than the origin because you will get (0,0,0 ) and the dot product cannot give you the answer you are looking for.

Now the parametric equations of the line in which the Jet fighter is lying are : x = 2+6m ,y = 8-12m , z = 1+12m
If P is the point on the line whose distance |OP| IS THE SHORTEST from the origin 0(0,0,0) then this distance is the magnitude of the vector
|vector OP| = sqroot((2+6m)^2+(8-12m)^2+(1+12m)^2) .
and as you may imagine this is a function of m!!
Get the first derivative of this function and find the stationary point of this function which expresses the magnitudes of all vectors starting from the origin and terminating at any point on the line. the value m=2/9 gives you where the stationary point is lying. and of course this st. point is a minimum
because if you find the second derivative and substitute there m=2/9 you will get an expression >0 .
Conclusion for m=2/9 you have the shortest distance |OP| of the line you are looking for from the origin.
The posision vector of the point P ( if my calculations are correct..... pls verify as well ) is (10/3)i+(16/3)j +(11/3)k .

MINOAS

Thankyou for the help!! ^

Yes, that is the answer. As of yet, we haven't done calculus, but I have a basic understanding from last year and you answer is informative. Thanks very much!

Another solution is to get the dot product between the vectors (2 + 6m)i + (8 - 12m)j + (1 + 12m)k and 6i - 12j + 12k) put them equal to zero and solve for m

you will find again m =2/9
the rest is easy...
good luck
Minoas

So would m = 2/9 then be the asymptote of the first derivative? (when you graph it)

MINOANMAN, I know how stupid I must seem, but can you exaplain why we'd find the dot product between (2 + 6m)i + (8 - 12m)j + (1 + 12m)k and (6i - 12j + 12k)? How does it show the shortest distance to the tower?

m=2/9 is the point (x-coordinate) that the first derivative is = zero because of the st. point.
do not confuse it with the asymptotes.....the asymptotes is another story.
By the way are ou an IG or a S.A.T student?

the shosrtest distance as chiro said " The smallest distance between two straight lines is at the points where the lines are perpendicular to each other"
therefore one vector is on the line and the other vector we want it to be perpendicular to it...thats why we get the dot product equal to zero...
it is easy
Minoas

Hahaha, I'm in Australia so I'm a 'VCE' student. Last year of school, highest level of maths available but the difficulty or structure should be similar to international courses I assume.
Thanks, that clears it up. :)