The shortest distance from a point to a plane is the length of the perpendicular line segment from the point to the plane. This distance represents the minimum separation between the point and any point on the plane.
To find the shortest distance from a point to a plane, we follow these steps:
The formula for the distance \(d\) from a point \((x_0, y_0, z_0)\) to a plane \(Ax + By + Cz + D = 0\) is:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Where:
Let's find the distance from the point P(1, 2, 3) to the plane 2x - y + 2z - 4 = 0:
Step 1: Identify A = 2, B = -1, C = 2, D = -4, x₀ = 1, y₀ = 2, z₀ = 3
Step 2: \(|Ax_0 + By_0 + Cz_0 + D| = |2(1) + (-1)(2) + 2(3) + (-4)| = |2 - 2 + 6 - 4| = |2| = 2\)
Step 3: \(\sqrt{A^2 + B^2 + C^2} = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3\)
Step 4: \(d = \frac{2}{3} \approx 0.667\)
In this diagram, the blue plane represents 2x - y + 2z - 4 = 0. The red point P(1,2,3) is shown above the plane. The red line segment represents the shortest distance from P to the plane, which we calculated to be approximately 0.667 units.
1. Distance to the xy-plane: For the xy-plane (z = 0), the formula simplifies to \(d = |z_0|\), where z₀ is the z-coordinate of the point.
2. Distance from the origin: For the origin (0, 0, 0), the formula becomes \(d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}\), where D is the constant term in the plane equation.
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