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Parameterizing Planes Question #2

  • #1
RJLiberator
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Homework Statement


Equation of the plane
Containing the line r(t) = <2 − t, 3, 4 + 2t> and point P(0, 0, 1).

Homework Equations


ax+by+cz=d equation of a plane

The Attempt at a Solution



1) We have a point, we need a normal vector
2) this time we are giving a line ON the plane, so the direction vector won't work as it is parallel to the plane.
3) Therefore, we can find the normal vector by using the point (0,0,1) and the point from the giving line (2, 3, 4) and making their line segment which comes out to be <-2, -3, -3>.
We then use this line segment and the direction vector <-1, 0, 2> take their cross product to net a vector that is perpendicular to the plane <6, -7, 3>.

And the equation of this plane becomes 6x-7y+3z=3
which satisfies both of our giving points (0, 0,1) and (2, 3, 4) on the plane.

The thing i am unsatisfied with here is why can we use 2 points on the plane, take their difference, and then cross it with the direction vector to get the perpendicular normal vector?

I understand that the cross product nets a perpendicular vector, but not the first part.
 

Answers and Replies

  • #2
Dick
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Homework Helper
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Homework Statement


Equation of the plane
Containing the line r(t) = <2 − t, 3, 4 + 2t> and point P(0, 0, 1).

Homework Equations


ax+by+cz=d equation of a plane

The Attempt at a Solution



1) We have a point, we need a normal vector
2) this time we are giving a line ON the plane, so the direction vector won't work as it is parallel to the plane.
3) Therefore, we can find the normal vector by using the point (0,0,1) and the point from the giving line (2, 3, 4) and making their line segment which comes out to be <-2, -3, -3>.
We then use this line segment and the direction vector <-1, 0, 2> take their cross product to net a vector that is perpendicular to the plane <6, -7, 3>.

And the equation of this plane becomes 6x-7y+3z=3
which satisfies both of our giving points (0, 0,1) and (2, 3, 4) on the plane.

The thing i am unsatisfied with here is why can we use 2 points on the plane, take their difference, and then cross it with the direction vector to get the perpendicular normal vector?

I understand that the cross product nets a perpendicular vector, but not the first part.
If you take two points in the plane and take their difference that gives you one vector that is tangent to the plane. The direction vector to the line is also tangent to the plane. If you take the cross product of the two that is perpendicular to both of them. Hence is a normal vector to the plane.
 
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  • #3
RJLiberator
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Okay, my followup question:

You say that the difference results in one vector that is tangent to the plane.
The direction vector is also tangent to the plane.

Tangent does NOT equal perpendicular, so we can't use these as normal vectors.
But does tangent = on the plane? If so, we can use these to take their cross product and receive a normal vector and all is well.
 
  • #4
Dick
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Okay, my followup question:

You say that the difference results in one vector that is tangent to the plane.
The direction vector is also tangent to the plane.

Tangent does NOT equal perpendicular, so we can't use these as normal vectors.
But does tangent = on the plane? If so, we can use these to take their cross product and receive a normal vector and all is well.
You are confusing position vectors with direction vectors. Only position vectors need to be 'on the plane'. The difference of two position vectors is a direction vector. The normal is not a position vector. It just points in a direction. Same for tangent. Is this making any sense?
 
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  • #5
RJLiberator
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Yes, getting there.

So a position vector comes from the origin to the point on the plane.
The direction vector does not do this, but instead can go through this 'line segment' that we've created from the difference.

So when taking the difference of two points on the plane we get a direction vector that 'connects' these two points. With this direction vector AND the giving direction vector of the line, this cross product results in the normal vector we are looking for.

Is that correct reasoning?
 
  • #6
Dick
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Yes, getting there.

So a position vector comes from the origin to the point on the plane.
The direction vector does not do this, but instead can go through this 'line segment' that we've created from the difference.

So when taking the difference of two points on the plane we get a direction vector that 'connects' these two points. With this direction vector AND the giving direction vector of the line, this cross product results in the normal vector we are looking for.

Is that correct reasoning?
Yes, some vectors indicate a position in space (position vectors). Some vectors just need to indicate a direction. The direction of <1,1,1> is the same as the direction of <2,2,2>. But regarded as position they are different.
 
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  • #7
RJLiberator
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Thank you for clarifying this for me. You have been a great advisor/helper. :)
 

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