Minimum Distance for Calculus

Question

A wandering albatross is sitting at the origin (0,0,0). What is the minimum distance between the albatross and a clothesline with equation r(vector)= [1,4,1] + s [-1,2,3] , s is every real number (seR).

Answer 1

Solution:

Method 1: Using vector algebra.

Let P = (1,4,1) and v=<-1,2,3>.

We first find the orthogonal projection of vector OP on vector v:

w = <1,4,1> ⋅ <-1,2,3> / |<-1,2,3>|^2 <-1,2,3> = 5/7 < -1,2,3>

Then the minimum distance is

d = |OP - w|

= |<1,4,1>-5/7<-1.2,3>|

= sqrt[(1+5/7)^2 + (4-10/7)^2 + (1-15/7)^2]

= 2/7 sqrt(133)

Answer: The minimum distance between the albatross and a clothesline is (2/7) sqrt(133).units.

Method 2: Using the derivative.

Let d be the distance between the origin and an arbitrary point (1-s, 4+2s, 1+3s) on the given line.

Then d^2 = f(s) = (1-s)^2 + (4+2s)^2 + (1+3s)^2

Now, f'(s) = -2(1-s) + 4(4+2s) + 6(1+3s).

Let f'(s) = 0 and solve for s to get the critical point s = -5/7.

According to the nature of the problem, d^2 = f(s) reaches the minimum value at s = -5/7:

f(-5/7) = (1+5/7)^2 + (4-10/7)^2 + (1-15/7)^2 = 532/47

Therefore, the minimum distance d = sqrt(532/47) = (2/7) sqrt(133).

Answer: The minimum distance between the albatross and a clothesline is (2/7) sqrt(133).units.