- #1

- 70

- 0

## Homework Statement

Find the distance between (2,5,1) and the line 2i − 3j + 6k.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter lockedup
- Start date

- #1

- 70

- 0

Find the distance between (2,5,1) and the line 2i − 3j + 6k.

- #2

- 65

- 1

... and the line 2i − 3j +6k.

It's a plane.

- #3

Mark44

Mentor

- 35,631

- 7,501

No, it isn't.It's a plane.

- #4

Mark44

Mentor

- 35,631

- 7,501

2i - 3j + 6k isn't a line -- it's a vector. It has a certain length, while a line has infinite length. The problem is probably something more like this:## Homework Statement

Find the distance between (2,5,1) and the line 2i − 3j + 6k.

Find the distance between (2,5,1) and the line whose direction is given by the vector 2i − 3j + 6k.

Just as well. Given that you can't find a formula, how would you approach this problem? According to the forum rules, you have to give it a good shot before anyone can give you any help.## Homework Equations

## The Attempt at a Solution

I can't find a formula to figure this (or one that makes any sense)...

- #5

- 9

- 0

I believe you are asking how to find the distance between a point in space, and a vector?

If so, start by looking at line-distance formulas and vector math.

I hope this gives you a jumping-off point.

- #6

- 70

- 0

My assignment sheet says line...2i - 3j + 6k isn't a line -- it's a vector. It has a certain length, while a line has infinite length. The problem is probably something more like this:

Find the distance between (2,5,1) and the line whose direction is given by the vector 2i − 3j + 6k.

Just as well. Given that you can't find a formula, how would you approach this problem? According to the forum rules, you have to give it a good shot before anyone can give you any help.

Does 20 or so google searches count? I've clicked on numerous links, some from here, and none of it makes any sense.

The formula in my Calculus book states:

[tex]D = \frac{||PQ \times u||}{||u||}[/tex]

P is a point on the line, Q is the point in space, and u is the direction vector. Since I'm only given a vector and as opposed to a line, can I use (0, 0, 0) for P so that PQ is just Q?

- #7

Mark44

Mentor

- 35,631

- 7,501

Sure, give your formula a shot.

And no, Google searches don't count...

And no, Google searches don't count...

- #8

- 65

- 1

No, it isn't.

You're right, I confused notation, sorry lockedup.

- #9

- 65

- 1

I believe you are asking how to find the distance between a point in space, and a vector?

If so, start by looking at line-distance formulas and vector math.

I hope this gives you a jumping-off point.

That was not my question.

- #10

Mark44

Mentor

- 35,631

- 7,501

Gunthi,

If it were 2x - 3y + 6z = 0, you would be right

If it were 2x - 3y + 6z = 0, you would be right

- #11

- 65

- 1

Gunthi,

If it were 2x - 3y + 6z = 0, you would be right

Yes, that was what I thought initialy.

I'm just not accostumed to working with i,j,k.

- #12

- 9

- 0

The formula wants you to multiply the scalar (point) by a unit vector and cross multiply with the given vector. Take the magnitude of the resultant vector. Then divide by magnitude of the unit vector (just a step that has to be done - balances things out . This will give a scalar quantity of distance.

Last edited:

Share: