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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1. How do you calculate the distance from a point to a plane?

The distance from a point to a plane can be calculated using the formula d = |ax + by + cz + d| / √(a^2 + b^2 + c^2), where (x,y,z) is the coordinates of the point and a, b, and c are the coefficients of the plane's equation (ax + by + cz + d = 0).

2. What is the significance of the absolute value in the distance formula?

The absolute value in the distance formula ensures that the distance is always positive, regardless of the sign of the result of the equation ax + by + cz + d. This is important because distance is a scalar quantity and cannot be negative.

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

No, the distance from a point to a plane cannot be negative. As mentioned earlier, distance is a scalar quantity and cannot have a negative value. The absolute value in the distance formula ensures that the result is always positive.

4. What are the units of measurement for the distance from a point to a plane?

The units of measurement for the distance from a point to a plane will depend on the units used for the coordinates of the point and the coefficients of the plane's equation. For example, if the coordinates are in meters and the coefficients are dimensionless, then the distance will be in meters.

5. How is the distance from a point to a plane related to the concept of perpendicularity?

The distance from a point to a plane is the shortest distance between the point and the plane. This distance is achieved when the line connecting the point to the plane is perpendicular to the plane. Therefore, the concept of perpendicularity is closely related to the calculation of the distance from a point to a plane.

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