calculate the distance from a point to a line in R3

Calculate the distance from P(1,1,1) and L: (0,-5/3,-8/3) + (lambda)(1,1,3)

Ended up with 1/33 sqrt(29^2 +(-26)^2 +(-1)^2). Have ended up with a very weird number!

Reply 1

DFranklin

I think you've made an arithmetic mistake somewhere. Also, if I'm not mistaken, the numbers have been chosen so it comes out reasonably nicely.

I suggest you post your working.

Reply 2

ghostwalker

Bit too slow:

I suspect a sign error - did the same thing myself first time.

I get

\sqrt{2}

, now.

Reply 3

Lunu

Original post by ghostwalker

Bit too slow:

I suspect a sign error - did the same thing myself first time.

I get

\sqrt{2}

, now.

Where was your sign error?

Reply 4

ghostwalker

Original post by Lunu

Where was your sign error?

Subtracting a negative dot product.

As DFranklin suggested, post your working.

Reply 5

Lunu

Original post by ghostwalker

Subtracting a negative dot product.

As DFranklin suggested, post your working.

I hope this attaches

12167828_10208097655987223_631109494_n.jpg55.1KB

Reply 6

ghostwalker

Original post by Lunu

I hope this attaches

You've taken the dot product of a with p, rather than with b.

Reply 7

Lunu

Original post by ghostwalker

You've taken the dot product of a with p, rather than with b.

So which two vectors should I use? Very confused!

Reply 8

ghostwalker

Original post by Lunu

So which two vectors should I use? Very confused!

PX, which is "b" in your working - the vector from the point P to a point on the line - and
"a" in your working - the direction vector for the line.

Edit: You worked out b, but then didn't use it.

(edited 8 years ago)

Reply 9

Lunu

Original post by ghostwalker

PX, which is "b" in your working - the vector from the point P to a point on the line - and
"a" in your working - the direction vector for the line.

Edit: You worked out b, but then didn't use it.