We've been going over the distance from point to plane. Its not too complicated although when we calculate the distance say from plan 2x+4y-4z-6=0 to point (0,3,6) you dot product the two vectors and take the absolute divided by the magnitude of the normal vector found in the plane equation. I'm wondering why in the top part of the solution, the absolute value part, why it is 2(0)+4(3)+6(-4)-6. Why do we include the -6? Wouldn't the normal vector be just <2,4,-4> multiplied by <0,3,6>? Please respond soon I have a test today.
Distance from point to plane
- Thread starter bearon37
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It appears that the way you've been taught to find the distance between a point and a plane is slightly different than what I learned. The way I learned was to find the normal vector of the plane \( \left( \vec{N} \right) \) and a vector from the plane to the point \( \left( \vec{W} \right) \). Then the distance is given by:
\(\displaystyle D = \frac{\biggr|\vec{N} \cdot \vec{W}\biggr|}{\biggr|\vec{N}\biggr|}\)
For this problem, we would have:
\(\displaystyle \vec{N} = \begin{bmatrix} 2 \\ 4 \\ -4 \end{bmatrix}\) and \(\displaystyle \vec{W} = \begin{bmatrix} x - 0 \\ y - 3 \\ z - 6 \end{bmatrix}\)
Plugging these in would give:
\(\displaystyle D = \frac{|2x + 4(y - 3) - 4(z - 6)|}{\sqrt{2^2 + 4^2 + (-4)^2}}\)
\(\displaystyle D = \frac{|2x + 4y - 4z + 12|}{6}\)
And by rearranging the given equation of the plane, we know that \(2x + 4y - 4z = 6\), so we can make that substitution:
\(\displaystyle D = \frac{6 + 12}{6} = \frac{18}{6} = 3\)
\(\displaystyle D = \frac{\biggr|\vec{N} \cdot \vec{W}\biggr|}{\biggr|\vec{N}\biggr|}\)
For this problem, we would have:
\(\displaystyle \vec{N} = \begin{bmatrix} 2 \\ 4 \\ -4 \end{bmatrix}\) and \(\displaystyle \vec{W} = \begin{bmatrix} x - 0 \\ y - 3 \\ z - 6 \end{bmatrix}\)
Plugging these in would give:
\(\displaystyle D = \frac{|2x + 4(y - 3) - 4(z - 6)|}{\sqrt{2^2 + 4^2 + (-4)^2}}\)
\(\displaystyle D = \frac{|2x + 4y - 4z + 12|}{6}\)
And by rearranging the given equation of the plane, we know that \(2x + 4y - 4z = 6\), so we can make that substitution:
\(\displaystyle D = \frac{6 + 12}{6} = \frac{18}{6} = 3\)
Wow thank you so much. That makes perfect sense. My professor is insane. He's the worst rated at my school on ratemyprofessor.
Dr.Peterson
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The method ksdhart2 demonstrated can be manipulated to obtain the formula your source used (and presumably stated somewhere). Searching for a place that teaches this formula, I found these which also derive it:
https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_plane, very last equation on page
https://mathinsight.org/distance_point_plane, again the last equation
It will be helpful if you show what you were taught, rather than just the example.
Dr.Peterson
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No one said that what you were taught was wrong, much less "insane" or "bad teaching". As far as I can tell so far, the error is on your part.
As I read it, you were taught a method involving a dot product, but you used the wrong vectors when you did that. But the solution to this problem that you saw (was it from your teacher, or somewhere else?) used the formula [MATH]{\dfrac {\left|ax_{1}+by_{1}+cz_{1}-d\right|}{\sqrt {a^{2}+b^{2}+c^{2}}}}[/MATH], which is perfectly valid.
I do hope you will show us the formula you were taught, so we can better explain how to apply that.
pka
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I fully understand Prof. Peterson's hope that bearon37 will post the method s/he was taught. Nevertheless. here is a way that some of us have found to be very clear. Lets use on your question.: the distance say from plane \(\displaystyle 2x+4y-4z-6=0\) to point \(\displaystyle P: (0,3,6)\).
Step:
1) write the normal: \(\displaystyle \vec{n}=<2,4,-4>\).
2) establish an 'easily found' point on the plane: \(\displaystyle A: (3,0,0)\)
3) write the vector: \(\displaystyle \overrightarrow {AP} = \left\langle { - 3,3,6} \right\rangle\)
4) apply the distance formula: \(\displaystyle \frac{|\overrightarrow {AP}\cdot\vec{n}|}{\|\vec{n}\|}=\frac{|-18|}{6}=3\)
HallsofIvy
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I would have done this is a slightly different way than pka:
The shortest distance from a given point to a given plane lies along the line through the given point perpendicular to the given plane.
1. write the normal: n→=<2,4,−4>. The same as pka
2. Write parametric equations for the line with direction vector <2, 4, -4> through P(0, 3, 6): x= 2t, y= 4t+ 3, z= -4t+ 6.
3. Find the point where that line passes through the plane by setting x, y, z in the equation of the plane: 2x+ 4y- 4z- 6= 2(2t)+ 4(4t+3)- 4(-4t+ 6)- 6= 36t- 18= 0. 36t= 18 so t= 18/36= 1/2. x= 2(1/2)= 1, y= 4(1/2)+ 3= 5, z= -4(1/2)+ 6= 4. The line passes through the plane at (1, 5, 4).
4. Find the distance between (0, 3, 6) and (1, 5, 4): sqrt((0-1)^2+ (3- 5)^2+ (6-4)^2)= sqrt(1+ 4+ 4)= 3.
The shortest distance from a given point to a given plane lies along the line through the given point perpendicular to the given plane.
1. write the normal: n→=<2,4,−4>. The same as pka
2. Write parametric equations for the line with direction vector <2, 4, -4> through P(0, 3, 6): x= 2t, y= 4t+ 3, z= -4t+ 6.
3. Find the point where that line passes through the plane by setting x, y, z in the equation of the plane: 2x+ 4y- 4z- 6= 2(2t)+ 4(4t+3)- 4(-4t+ 6)- 6= 36t- 18= 0. 36t= 18 so t= 18/36= 1/2. x= 2(1/2)= 1, y= 4(1/2)+ 3= 5, z= -4(1/2)+ 6= 4. The line passes through the plane at (1, 5, 4).
4. Find the distance between (0, 3, 6) and (1, 5, 4): sqrt((0-1)^2+ (3- 5)^2+ (6-4)^2)= sqrt(1+ 4+ 4)= 3.