# What is the distance of closest approach

#### skeeter

##### Well-known member
MHB Math Helper
If $P$ is a point in space, and $L$ is the line $r(t) = \vec{r_0} + \vec{A} t$ , then

$d(P,L) = \dfrac{|\vec{Pr_0} \times \vec{A} |}{|\vec{A}|}$

• Another

#### Country Boy

##### Well-known member
MHB Math Helper
View attachment 10653

from problem I find $r = r_0 + At$ $x_0 = 3 + 2t$ $y_0 = -1 - 2t$ $z_0 = 1 + t$ and $A = (2,-2,1)$
you haven't actually written the path in the form r= r0+ At. It is, of course, r= <3, -1, 1>+ <2, -2, 1>t.

but i don't understand What is the distance of closest approach?

someone tell me to a formula please.
It is the shortest possible distance from a point, (x(t), y(t), z(r)), on the graph to the origin, (0, 0, 0). The "formula" you want is the distance formula. The distance from (x, y, z) to the origin is $\sqrt{x^2+ y^2+ z^2}$. "Minimizing" that distance is the same as minimizing the square- $x^2+ y^2+ z^2= (3+ 2t)^2+ (-1- 2t)^2+ (1+ t)^2= 9+ 12t+ 4t^2+ 1+ 4t+ 4t^2+ 1+ 2t+ t^2= 11+ 18t+ 10t^2$. You can minimize that by setting the derivative equal to 0 or by "completing the square".

To complete the square, \$10t^2+ 18t+ 11= 10(t^2+ 1.8t)+ 11= 10(t^2+ 1.8t+ 0.81- 0.81)+ 11= 10(t+ .9)^2- 8.1+ 11= 10(t+ .9)^2+ 2.9. Since a square is never negative, that is smallest when t= -0.9 where its value is 2.9.

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