Equation of plane through point P containing line L

In summary, the equation for the plane W2 through the point P(2,3,-1) containing L is 14x - 7y + 21z + 14 = 0, with two points on L being Q(8,-2,3) and R(13,-3,6). The normal to the plane W2 is given by the cross product of vectors PQ and PR, which is (14,-7,21). It is important to carefully check calculations and reasoning to ensure the accuracy of mathematical solutions.
  • #1
phyzmatix
313
0

Homework Statement



Find an equation for the plane W2 through the point P(2,3,-1) containing L

[L = (3,-1,0) + t(5/3, -1/3, 1)]2. The attempt at a solution

Let Q and R be two points on L where t = 3 and t = 6 respectively

(t = 3) Q(11, -3, 3)
(t = 6) R(16, -4, 6)

Then PQ = (9, -6, 4) and PR = (14, -7, 7)

The normal to the plane W2 is then

PQ x PR = (14, -7, 21)

Finally

W2 : a(x - x0) + b(y - y0) + c(z - z0) = 0
14(x - 2) - 7(y - 3) + 21(z - +1) = 0
14x -7y +21z + 14 = 0

Could someone please be so kind as to check my calculations (especially for the two points Q and R, I'm not sure if I derived them correctly from the line's equation)? I'm not sure if my answer is correct since I can't find any similar examples in my textbook so I'd just like to know if my reasoning is sound.

Thanks in advance.

phyz
 
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  • #2
phyzmatix said:
L = (3,-1,0) + t(5/3, -1/3, 1) ...Let Q and R be two points on L where t = 3 and t = 6 respectively

(t = 3) Q(11, -3, 3)
(t = 6) R(16, -4, 6)
I understand why you chose t = 3 and t = 6 (though why not use t = 0?). But the values should be

Q: (3,-1,0) + 3(5/3, -1/3, 1) = (3+5, -1-1, 0+3) = (8, -2, 3) and
R: (3,-1,0) + 6(5/3, -1/3, 1) = (3+10, -1-2, 0+6) = (13, -3, 6) ,

shouldn't they?

Otherwise, in principle, your method looks all right...
 
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  • #3
dynamicsolo said:
I understand why you chose t = 3 and t = 6 (though why not use t = 0?). But the values should be

Q: (3,-1,0) + 3(5/3, -1/3, 1) = (3+5, -1-1, 0+3) = (8, -2, 3) and
R: (3,-1,0) + 6(5/3, -1/3, 1) = (3+10, -1-2, 0+6) = (13, -3, 6) ,

shouldn't they?

Otherwise, in principle, your method looks all right...

Thanks dynamicsolo! I just had a look here and I honestly can't tell you how I got to those values :confused:

Just goes to show that you shouldn't attempt mathematics at the end of the day when the old grey matter is starting to rebel against work :biggrin:
 
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  • #4
phyzmatix said:
Just goes to show that you shouldn't attempt mathematics at the end of the day when the old grey matter is starting to rebel against work :biggrin:

I tell the students I work with that mathematical reasoning engages many of the so-called "higher brain functions" (there is also a lot of interaction going on between various sections of the brain). Those functions are the first thing to go when you're tired or hungry, so you become prone to increasing numbers of mistakes (sometimes bizarre ones -- I've written stuff at 2 AM that made next to no sense after a night's sleep...).
 
  • #5
I am facing a similar problem right now, and i understand most of it, however, i am just having a hard time seeing how PQ * PR = normal of the plane
Thank you!
 
  • #6
cemar. said:
I am facing a similar problem right now, and i understand most of it, however, i am just having a hard time seeing how PQ * PR = normal of the plane
Thank you!
What do you mean by "PQ*PR"? It is the cross product of two vectors that is perpendicular to the plane containing the vectors and the standard notation for cross product is [itex]PQ \times PR[/itex]. And, of course, it is normal to the plane because the two vectors are in the plane and, as I said, "the cross product of two vectors is perpendicular to the plane containing them".
 

1. What is an equation of a plane?

An equation of a plane is a mathematical expression that represents a flat surface in three-dimensional space. It is typically written in the form of Ax + By + Cz + D = 0, where A, B, and C are the coefficients of the variables x, y, and z, and D is a constant.

2. How do you find the equation of a plane through a given point?

To find the equation of a plane through a given point, you need to know the coordinates of the point and the normal vector of the plane. The equation can then be written as (x-x0)n1 + (y-y0)n2 + (z-z0)n3 = 0, where (x0, y0, z0) are the coordinates of the given point and (n1, n2, n3) is the normal vector.

3. What is a normal vector of a plane?

A normal vector of a plane is a vector that is perpendicular to the plane. It determines the direction of the plane's surface and is essential in writing the equation of the plane.

4. How do you find the equation of a plane containing a given line?

To find the equation of a plane containing a given line, you need to know the coordinates of a point on the line and the direction of the line. The equation can then be written as (x-x0)n1 + (y-y0)n2 + (z-z0)n3 = 0, where (x0, y0, z0) are the coordinates of the given point on the line and (n1, n2, n3) is the direction vector of the line.

5. Can a plane contain more than one line?

Yes, a plane can contain an infinite number of lines. In fact, any three non-collinear points determine a unique plane, and any line passing through those points will lie on that plane.

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