Calculating the distance from a point to a plane

In summary, the conversation discusses finding the distance between a point and a plane using the formula d=|PQ*n| and choosing a point on the plane. There is confusion about how to determine if a point is on the plane and whether a parallelogram can be formed without the point touching the plane. The solution provided has some errors and another approach is suggested.
  • #1
Darkmisc
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Homework Statement
What is the distance from r*(2i -j -2k) = 7 to the point (1, 1, -1)?
Relevant Equations
d=|PQ*n|
Hi everyone

I have worked solutions to the question, but I don't fully understand what they are doing and I get a different answer.

I used d=|PQ*n| and chose (0, 0, -7/2) as a point on the plane. I got that point by letting i and j = 0.

Since P = (1, 1, -1), PQ = (-1, -1, -5/2).

The unit vector of the normal works out to be (1/3)(2i -j -2k)

So d = |(-1, -1, -5/2)*(1/3)*(2, -1, -2)| = 4/3.

These are the worked solutions.
1679531092451.png


I understand the general logic behind it and can accept that OB' = BM. However, I don't understand how they know M is a point on the plane. Is it not possible for OB'MB to form a parallelogram without M touching the plane?Thanks
 
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  • #2
Darkmisc said:
I understand the general logic behind it and can accept that OB' = BM. However, I don't understand how they know M is a point on the plane. Is it not possible for OB'MB to form a parallelogram without M touching the plane?
You define M to be the point on the plane closest to B.

That said, the solution is full of mistakes. First, it should be ##\overrightarrow{OA} = \frac{7}{2} \mathbf{k}##.

Then, I get
$$
\frac{\overrightarrow{OB} \cdot \mathbf{n}}{|\mathbf{n}|} = 1,
$$
not ##1/3##, such that ##|\overrightarrow{OB'}| = 1##. Since ##|\overrightarrow{ON}| = 7/3##, this means that ##|\overrightarrow{B'N}| = 4/3##. I then take ##|\overrightarrow{BM}| = |\overrightarrow{B'N}| = 4/3##, so I get the same answer as you did (and I like your approach better). I'm not sure I get the parallelogram either.
 
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FAQ: Calculating the distance from a point to a plane

What is the formula to calculate the distance from a point to a plane?

The formula to calculate the distance from a point \((x_1, y_1, z_1)\) to a plane defined by \(Ax + By + Cz + D = 0\) is given by:\[ \text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

How do you derive the distance formula from a point to a plane?

The distance formula is derived using the concept of the perpendicular (normal) distance from the point to the plane. By projecting the point onto the plane along the normal vector of the plane, the distance can be calculated as the absolute value of the dot product of the point's position vector with the plane's normal vector, divided by the magnitude of the normal vector.

What do the variables A, B, C, and D represent in the plane equation?

In the plane equation \(Ax + By + Cz + D = 0\), \(A\), \(B\), and \(C\) are the coefficients that define the normal vector to the plane, and \(D\) is the constant term that shifts the plane relative to the origin. The normal vector \((A, B, C)\) is perpendicular to the plane.

Can the distance from a point to a plane be negative?

No, the distance from a point to a plane cannot be negative because distance is always a non-negative scalar quantity. The absolute value in the formula ensures that the distance is positive or zero (if the point lies on the plane).

How do you interpret the result if the distance is zero?

If the calculated distance is zero, it means that the point lies exactly on the plane. This occurs when the point satisfies the plane equation \(Ax + By + Cz + D = 0\).

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