Distance from a point to a plane.

In summary, the conversation discusses finding the distance of a point from a plane determined by three other points, using common vectors and the normal vector to the plane. The final calculation involves a dot product and division by the magnitude of the normal vector. The answer may have a typo in the denominator.
  • #1
Gregg
459
0

Homework Statement



Find the distance of the point (2,4,7) from the plane determined by the points (3,4,4) , (4,5,7) and (6,8,9).

2 common vectors for the plane

[itex]\left(
\begin{array}{c}
1 \\
1 \\
3
\end{array}
\right)
[/itex] [itex] \left(
\begin{array}{c}
3 \\
4 \\
5
\end{array}
\right) [/itex]

The normal to the plane is then

[itex]\left(
\begin{array}{c}
1 \\
1 \\
3
\end{array}
\right)\times \left(
\begin{array}{c}
3 \\
4 \\
5
\end{array}
\right)=\left(
\begin{array}{c}
-7 \\
4 \\
1
\end{array}
\right)[/itex]

The vector from the point to a known point on the plane dot product with the normal vector divided by |n| gives

[itex]\frac{\left(
\begin{array}{c}
1 \\
0 \\
-3
\end{array}
\right).\left(
\begin{array}{c}
-7 \\
4 \\
1
\end{array}
\right)}{\sqrt{7^2+4^2+1^2}}=\frac{10}{\sqrt{66}} = \frac{5\sqrt{66}}{33}[/itex]

The answer has just 3 in the denominator, not 33. Have I done this correctly?
 
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  • #2
That looks fine to me. May be a typo in the answers.
 
  • #3


I would first like to commend you for your efforts in solving this problem. Your approach is correct, however, there may be a minor error in your calculations. When taking the dot product of the vector from the point to a known point on the plane and the normal vector, you should have 10 as the numerator and 66 as the denominator, giving you a final answer of 5/33. It seems that you may have mistakenly squared the components of the normal vector when calculating the magnitude. I would recommend double-checking your calculations to ensure accuracy. Overall, your method is sound and your solution is correct. Great job!
 

What is the formula for finding the distance from a point to a plane?

The formula for finding the distance from a point to a plane is d = |Ax + By + Cz + D| / √(A² + B² + C²), where (x, y, z) is the coordinates of the point and A, B, C, and D are the coefficients of the plane's equation in the form Ax + By + Cz + D = 0.

How do you determine if a point is above or below a plane?

To determine if a point (x, y, z) is above or below a plane, you can plug in the coordinates into the plane's equation Ax + By + Cz + D and check if the resulting value is positive or negative. If it is positive, the point is above the plane, and if it is negative, the point is below the plane.

What is the significance of the distance from a point to a plane?

The distance from a point to a plane is the shortest distance between the point and the plane. It is useful in many applications, such as determining the closest point on a plane to a given point, finding the intersection of a line and a plane, and calculating the angle between a line and a plane.

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

No, the distance from a point to a plane cannot be negative. It is always a positive value, as it is the measure of the shortest distance between a point and a plane, which is always a straight line.

How does the distance from a point to a plane change if the plane is rotated or translated?

The distance from a point to a plane remains the same regardless of the plane's rotation or translation. This is because the formula for distance is based on the plane's equation, which remains unchanged even after rotation or translation. Only the coordinates of the point may change, but the distance remains constant.

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