Question

Find the distance from the point (-3,2,1) to the plane 1x+5y-2z=-3

Answer 1

find a set of parametric equations for the line through the point that is perpendicular to the plane, then find the intersection of the line and the plane and plug this into the parametric equation

Answer 2

You want to find any vector that goes from the point to the plane. In this case it is <-3-x, 2-y, 1-z>. Since the normal vector is be the shortest distance, you want to project onto that normal vector. The normal vector is <1, 5, -2> in this case. The magnitude of that projection is the shorted distance from the plane to the point.

Do you know how to project a vector?

Answer 3

Yeah I got the formula around here somewhere

Answer 4

wio can you help me out, because im not really sure what im doing

Answer 5

@math_proof with what?

Answer 6

with that proof

Answer 7

(u*v/||u^2||)*u

Answer 8

that i posted

Answer 9

ok

Answer 10

http://mathinsight.org/distance_point_plane

Answer 11

@ConDawg, still need help?

Answer 12

Yea, so if Normal vector is <1,5,-2>, and wwe want to project it on <-3,2,1>?

Answer 13

We want to project <-3-x, 2-y, 1-z> onto it.

Answer 14

And we want to find the magnitude of that projection.

Answer 15

So we take the dot product of <-3-x, 2-y, 1-z> and <1,5,-2> first of all.

Answer 16

Then we divide that by |<1,5,-2>|, the magnitude of the normal vector.

Answer 17

So 5/sqrt(30)

Answer 18

No, the dot product should have x, y, and z in it
Where did they go?

Answer 19

i was taught to always give a new points not the magnitude of it

Answer 20

We'll get rid of them soon enough, but unless you're ahead of the game, they should still be in there.

Answer 21

@ConDawg What is the dot product of <-3-x, 2-y, 1-z> and <1,5,-2>?

Answer 22

The dot product gives a scaler, which is just a single number, is what my text book says, so should I ignore that part for now?

Answer 23

We will get a scalar, it's just that it should have x, y, and z in it.

Answer 24

wio can you give me some guide to do part b of my proof?

Answer 25

I get 1(-3-x) + 5(2-y) - 2(1-x)

Answer 26

Okay

Answer 27

Now when you expand that you get?

Answer 28

-3-x+10-5y-2+2x

Answer 29

Okay, all the like terms need to be added up...

Answer 30

Like -3, 10, and -2 are like terms.

Answer 31

-x-5y+2x+5

Answer 32

Now we use the fact that 1x+5y-2z=-3 to get rid of our x, y and z.

Answer 33

If 1x+5y-2z=-3 then -x-5y+2z = 3

Answer 34

Okay got cha

Answer 35

So it's \(\Large \frac{3+5}{\sqrt{30}} or \frac{8\sqrt{30}}{30} \)

Answer 36

I hate this stuff

Answer 37

LOL, me too!

Answer 38

Nice its right thanks a bunch!!!!