Find the Vector Equation of Line Perp. to r Passing Through P(1,3,5)

[adrimare]

Homework Statement

Determine the vector equation of the line that passes through the point P(1,3,5) and is perpendicular to the line r = (1,0,5) + t(-3,4,-6).

Homework Equations

n1 dot n2 = 0

The Attempt at a Solution

I know that the normal of the line I need to find and the normal of the line perpendicular to it have to dot to equal 0. The direction vectors of both lines also can dot to equal 0, too, right? I'm just confused on how to get the needed line and/or the normals of both lines.

 

[Mark44]

Homework Statement

Determine the vector equation of the line that passes through the point P(1,3,5) and is perpendicular to the line r = (1,0,5) + t(-3,4,-6).

Homework Equations

n1 dot n2 = 0

The Attempt at a Solution

I know that the normal of the line I need to find and the normal of the line perpendicular to it have to dot to equal 0. The direction vectors of both lines also can dot to equal 0, too, right? I'm just confused on how to get the needed line and/or the normals of both lines.

This is similar to another problem you posted about two parallel lines. In this problem you need to find the dot product of the line's direction vector with a vector from (1, 3, 5) to the line. Every point on the line can be expressed as (1 - 3t, 4t, 5 - 6t).

 

[adrimare]

So I did (1 - 3t, 4t, 5 - 6t) - (1,3,5) and got (3t,4t - 3,6t). So I dotted that by (-3,4,-6) and made it equal to 0. I got -9t + 16t - 12 - 36t = 0. This becomes -29t = 12, which means that t = -29/12. The answer is supposed to be (0,6,4) for the direction vector of the line I want to find. What am I doing wrong? Or is the book answer section wrong again?

 

[Mark44]

So I did (1 - 3t, 4t, 5 - 6t) - (1,3,5) and got (3t,4t - 3,6t). So I dotted that by (-3,4,-6) and made it equal to 0. I got -9t + 16t - 12 - 36t = 0. This becomes -29t = 12, which means that t = -29/12. The answer is supposed to be (0,6,4) for the direction vector of the line I want to find. What am I doing wrong? Or is the book answer section wrong again?

The purpose of finding the value of t was not just to find a value of t -- it was to find the point on the line. Find the coordinates of the point by substituting the value of t you found in (1 - 3t, 4t, 5 - 6t). BTW, I haven't checked your work for your value of t, so can't guarantee that you'll get the right point.

For one thing, (1 - 3t, 4t, 5 - 6t) - (1,3,5) != (3t, 4t - 3, 6t). Check your arithmetic.

 

[adrimare]

That's what I got, too. Also, I made a mistake with the t-value. It was -12/29. Either way, it doesn't get me (0,6,4), which is the direction vector in the answer section of the book. Am I doing something wrong or is the book wrong?

 

[Mark44]

Are you sure you wrote the problem correctly? I have looked a little more closely at your problem, as you stated it.

Determine the vector equation of the line that passes through the point P(1,3,5) and is perpendicular to the line r = (1,0,5) + t(-3,4,-6).

There are an infinite number of lines that are perpendicular to the given line and that pass through P(1, 3, 5). They form the plane whose equation is 3x - 4y + 6z = 21. Notice that <3, -4, 6> is a normal to the plane, and that this vector is paralles to the line's direction (but points the opposite direction). Note also that P(1, 3, 5) satisfies the equation of the plane, since 3(1) - 4(3) + 6(5) = 21.

 

[adrimare]

How did you get the = 21 part of the Cartesian equation? The point (1,0,5) does not work with that. Is it just a part of the line that is not on the plane or something? Would this plane look like a cube? The question also asks for parametric and symmetric equations, but since you can get those from the vector equation, I did not include that part. Anyways, the vector equation is the one I should be finding here primarily because you can use a point and a direction vector to make it.

Also, why do the two lines make a plane with that normal? The line (x,y,z) = (1,0,5) + t(-3,4,-6) is perpendicular to a plane with the normal of (3,-4,6), right? But a line perpendicular to it would have a different direction vector, and the plane formed by them would have a vector of the cross product of the two direction vectors, right? I'm just looking for that direction vector so that I can complete the vector equation needed of r = (1,3,5) + t(a,b,c).

 

[Mark44]

You didn't respond to my question about whether you posted the problem correctly.

How did you get the = 21 part of the Cartesian equation? The point (1,0,5) does not work with that. Is it just a part of the line that is not on the plane or something? Would this plane look like a cube? The question also asks for parametric and symmetric equations, but since you can get those from the vector equation, I did not include that part. Anyways, the vector equation is the one I should be finding here primarily because you can use a point and a direction vector to make it.

The point (1, 0, 5) doesn't have anything to do with it, so doesn't need to satisfy the plane equation.

I got the equation I gave by making a vector from an arbitrary point (x, y, z) on the plane to (1, 3, 5), and dotting this vector with the direction of the line <-3, 4, -6>. Since the vector in the plane is perpendicular to the direction of the line, the dot product is 0.

No, a plane doesn't look like a cube. How could it? It's a two-dimensional flat surface.

Also, why do the two lines make a plane with that normal?

You are given only one line. There are an infinite number of lines that are perpendicular to the given line and pass through the given point. Those lines define and generate the plane whose equation I gave. Think of a string stretched tight in space with a point marked on it. Now put the eraser end of a pencil at the point and orient the pencil so that it is perpendicular to the string. You can swing the pencil around in a circle around the string so that in each position the pencil is perpendicular to the string. In each position, the direction vector of the pencil is different, but the dot product of the pencil's direction and the string's direction will be 0.

The line (x,y,z) = (1,0,5) + t(-3,4,-6) is perpendicular to a plane with the normal of (3,-4,6), right?

Right.

But a line perpendicular to it would have a different direction vector, and the plane formed by them would have a vector of the cross product of the two direction vectors, right? I'm just looking for that direction vector so that I can complete the vector equation needed of r = (1,3,5) + t(a,b,c).

 

[adrimare]

Yes, I did input the right question. I tried doing what you said, and it still isn't working. Oh well. Thanks.

 

[Mark44]

I guess you're talking about this:

I got the equation I gave by making a vector from an arbitrary point (x, y, z) on the plane to (1, 3, 5), and dotting this vector with the direction of the line <-3, 4, -6>. Since the vector in the plane is perpendicular to the direction of the line, the dot product is 0.

<x -1, y -3, z -5>.<-3, 4, -6> = 0
==> ??
Multiply the dot product. What equation do you get?

 

[adrimare]

That works. I got 3x - 4y + 6z -21 = 0. I meant that I tried the stuff in your previous posts and can't match the book answer. I can't get a direction vector to be (0,6,4) or (0,3,2). Because the t-values don't work.

 

[Mark44]

Those answers work as direction vectors, which you can verify by dotting them with <-3, 4, -6>, but I have no idea how the textbook author came up with them.

Any vector that is perpendicular to <-3, 4, -6> will work, which represents an infinite number of vectors, all of which lie in the plane 3x - 4y + 6z = 21.

Seems like a flaky problem to me.