# Vector geometry help

1. Jan 28, 2005

### mathrocks

I have a question about solving the following problem
"Find the line through (3,1,-2) that intersects and is perpendicular to the line x=-1+t, y=-2+t, z=-1+t."

My question is regarding the intersection part, I know how to find the equation of the line that is perpendicular to another line but not both perpendicular and intersecting.

thanks

2. Jan 28, 2005

### Phymath

I'd think your forgeting that the dot product of two perpendicular vectors is 0 think about it...

3. Jan 28, 2005

### vincentchan

what?
dot product....
the answer is as easy as you can think

(3,1,-2)+t(1,1,1)

why the hell you need dot product?

4. Jan 28, 2005

### daster

If two lines are perpendicular to each other, don't you think they're already intersecting?

5. Jan 28, 2005

### Muzza

And that line is perpendicular to x=-1+t, y=-2+t, z=-1+t?

No, two perpendicular lines in R^3 don't necessarily intersect.

Last edited: Jan 28, 2005
6. Jan 28, 2005

### mathrocks

Wouldn't (3,1,-2)+t(1,1,1) be correct for the line perpendicular to x=-1+t, y=-2+t, z=-1+t since the vector <1,1,1> is coming from that?

7. Jan 28, 2005

### Muzza

A direction vector of (3,1,-2)+t(1,1,1) is (1,1,1), and a direction vector of x=-1+t, y=-2+t, z=-1+t is (1,1,1). Two lines are perpendicular if and only if their direction vectors are perpendicular. Surely (1,1,1) is not perpendicular to (1,1,1)?

8. Jan 28, 2005

### mathrocks

So, how would you go about finding a line that is perpendicular?

9. Jan 28, 2005

### daster

Ah.. I over-looked that.

10. Jan 28, 2005

### daster

The dot product between their direction vectors should be 0.

11. Jan 28, 2005

### HallsofIvy

Staff Emeritus
Note: there exist an infinite number of lines (in 3D) through a given point and perpendicular to a given line. Perhaps part of the problem is that there are too many answer. You should know that, in 2D, a perpendicular to <a, b> is <b, -a> since that way <a, b>.<b,-a>= ab-ab= 0. In 3d, the simplest thing to do is just take one of the components 0 so that the problem goes back to 2D.

To find a perpendicular to <1, 1, 1> try <0, a, b> What must a, b be?
Of course, <a, 0, b> and <a, b, 0> would work as well.

12. Jan 28, 2005

### Phymath

the line has the vector <1,1,1> as its direction
dot with the vector that lies on the perp line c...
$$\vect{c} = <c_x, c_y, c_z>$$
$$<1,1,1> dot <c_x, c_y,c_z> = 0$$
gives you
$$c_x = -c_y - c_z$$
$$c_y free$$
$$c_z free$$

subracting the vector that points to the point on the line of perpendicularity (r), and the vector the point we are trying to make a line through gives us the vector in the direction of the line which is c...
$$<r_x, r_y, r_z> - <3,1,-2> = <-c_y-c_z, c_y, c_z>$$
$$r_x - 3 = -c_y - c_z$$
$$r_y - 1 = c_y$$
$$r_z + 2 = c_z$$

however the terminal point of vector r must lie on the origal line..there fore..

r_x = t -1
r_y = t - 2
r_z = t - 1 ...substute...

$$(t-1) - 3 = -c_y - c_z$$
$$(t-2)-1 = c_y$$
$$(t - 1) + 2 = c_z$$

which we can easily solve for t...t = 2 put find the vector c by substuting in the t value...you'll get...
$$\vect{c} = <-2, -1, 3>$$
now we can form the line equation
x = -2t +3
y = -t + 1
z = 3t -2

done...check the dot product its perp

Last edited: Jan 28, 2005
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