Question

.

Answer 1

errrrr... are you sure you can't plug the Point into the x y z?

Answer 2

oh distance formula...

Answer 3

point to the line distance formula use that.

Answer 4

you familiar wid vectors ?

Answer 5

say P' = <3, 0, 1>, then :

direction vector : <-1, 1, 2>
PP' = <5, -1, 0>

Answer 6

P' is just some point on the given line

Answer 7

once we have above, we can calculate the distance :
1) find the area of parallelogram formed by vectors : PP' and 'direction vector'
2) divide the area by base (magnitude of direction vector)

Answer 8

another brute-force way that might be a bit easier is to minimize the distance squared from point P to point (x(t),y(t),z(t)) on line L

Answer 9

area of parallelogram = PP' x direction vector
= <5, -1, 0> x <-1, 1, 2>

Answer 10

minimizing distance works too, as we wil be having all square terms in the distance formula :)

Answer 11

so, distance =

|<5, -1, 0> x <-1, 1, 2> |
----------------------
|<-1, 1, 2> |

Answer 12

good :) try minimizing the distance also... that involves oly algebra... u dont need calculus... thats the fastest way to find distance i feel

Answer 13

any point on the line is given by parameter :
(3-t, t, 1+2t)

Answer 14

given point : P(-2, 1, 1)

Answer 15

plug in the - 2 1 1 to (3-t, t, 1+2t)

Answer 16

5 1 3

Answer 17

$$
D=|<x,y,z> - <-2,1,1>|\\
|<x+2,y-1,z-1>|\\
=\left ((x+2)^2+(y-1)^2+(z-1)^2\right )^{1/2}\\
$$
Now substitute x=3-t, y=t,z=1+2t:
$$
D= \sqrt{6 t^2-12 t+26}
$$

Answer 18

consider minimizing D^2, a quadratic function (6 t^2-12 t+26)

Answer 19

D^2 is has a minimum of 20 at t=1; thus the minimum distance, sqrt(D^2), is sqrt(20)