To apply projection of linear algebra, it's more easy than derivative
For the point P = (1,2,3) and the line x = 8-2t, y = -5+t, z = -4 + 3t
Find the distance between P and an arbitrary point on the line, in terms of the parameter t.
i chose t = 0 so the arbitrary point Q = (8,-5,-4)
then,
PQ = <7,-7,-7>
so the distance would be
Does this look correct?
Then the next part of this question states:
Find the value of t that minimizes the distance function above
i know that to find a minimization you take the derivative of the function but im not sure what to take the derivative of in this case.... or if thats what i even need to do.
Thanks for any help
No, it is not! An "arbitrary point" means just that. Once you choose a point, it is not arbitrary! Did you see where the problem says "in terms of the parameter t"? What they mean is to find the distance between (1, 2, 3) and the "arbitrary point" (8- 2t, -5+ t, -4+ 3t).
No, it isn't. The distance between two point (x, y, z) and (a, b, c) is given by . What you have is the distance from your "arbitrary" point (7, -7, 7) and the origin. What happened to the point (1, 2, 3)?then,
PQ = <7,-7,-7>
so the distance would be
The distance between (1, 2, 3) and an arbitrary point on the line x = 8-2t, y = -5+t, z = -4 + 3t and (1, 2, 3) is given by .
What is that equal to?
You have completely misunderstood the question. See above.Does this look correct?
Yes, you do- but having chosen a specific point rather than taking an "arbitrary point" "in terms of t", you don't have function of t!Then the next part of this question states:
Find the value of t that minimizes the distance function above
i know that to find a minimization you take the derivative of the function but im not sure what to take the derivative of in this case.... or if thats what i even need to do.
Thanks for any help
Take the derivative of with respect to t.
(Hint: the derivative of is and is 0 if and only if f'(x) is 0.)