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Three-Dimensional Geometry - Distance of a Point from a Plane

Grade 12ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Perpendicular Distance Concept: The distance of a point P(x1,y1,z1)P(x_1, y_1, z_1) from a plane is defined as the length of the perpendicular segment dropped from the point to the plane. Visually, imagine a point floating in space and a flat sheet representing the plane; the shortest path from the point to the sheet is the straight line that meets the sheet at a 90∘90^{\circ} angle.

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Normal Vector Orientation: In the Cartesian equation of a plane Ax+By+Cz+D=0Ax + By + Cz + D = 0, the coefficients (A,B,C)(A, B, C) represent the direction ratios of the normal vector n⃗\vec{n}. This normal vector is geometrically a line protruding straight out of the plane surface, indicating its orientation in 3D space.

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Vector Form Perspective: For a plane given by the vector equation r⃗⋅n⃗=d\vec{r} \cdot \vec{n} = d, where n⃗\vec{n} is the normal vector and dd is a scalar, the distance from a point with position vector a⃗\vec{a} is determined by how much the vector (a⃗)(\vec{a}) deviates from the plane's boundary along the direction of the unit normal.

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Distance from the Origin: This is a specific case where the point is (0,0,0)(0, 0, 0). Geometrically, this represents the length of the normal drawn from the origin to the plane. If the plane is in the form lx+my+nz=plx + my + nz = p (where l,m,nl, m, n are direction cosines), then pp is the actual distance from the origin.

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Parallel Planes Distance: When two planes are parallel, they share the same normal direction (A,B,C)(A, B, C) but have different constants D1D_1 and D2D_2. Visually, they look like two parallel floors in a building, and the distance between them is constant regardless of where it is measured.

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Sign of the Distance: While distance is always a non-negative magnitude (indicated by absolute value bars ∣...∣|...|), the sign of the expression Ax1+By1+Cz1+DAx_1 + By_1 + Cz_1 + D indicates which side of the plane the point lies on relative to the origin.

📐Formulae

Cartesian Distance from P(x1,y1,z1)P(x_1, y_1, z_1) to Ax+By+Cz+D=0Ax + By + Cz + D = 0: d=∣Ax1+By1+Cz1+D∣A2+B2+C2d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}

Vector Distance from point a⃗\vec{a} to plane r⃗⋅n⃗=q\vec{r} \cdot \vec{n} = q: d=∣a⃗⋅n⃗−q∣∣n⃗∣d = \frac{|\vec{a} \cdot \vec{n} - q|}{|\vec{n}|}

Distance from Origin (0,0,0)(0, 0, 0) to Ax+By+Cz+D=0Ax + By + Cz + D = 0: d=∣D∣A2+B2+C2d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}

Distance between parallel planes Ax+By+Cz+D1=0Ax + By + Cz + D_1 = 0 and Ax+By+Cz+D2=0Ax + By + Cz + D_2 = 0: d=∣D1−D2∣A2+B2+C2d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}

Vector form with unit normal: If n^\hat{n} is a unit vector, the distance from a⃗\vec{a} to r⃗⋅n^=p\vec{r} \cdot \hat{n} = p is d=∣a⃗⋅n^−p∣d = |\vec{a} \cdot \hat{n} - p|

💡Examples

Problem 1:

Find the distance of the point (2,5,−3)(2, 5, -3) from the plane 6x−3y+2z−4=06x - 3y + 2z - 4 = 0.

Solution:

  1. Identify the coordinates of the point: x1=2,y1=5,z1=−3x_1 = 2, y_1 = 5, z_1 = -3.
  2. Identify coefficients from the plane equation: A=6,B=−3,C=2,D=−4A = 6, B = -3, C = 2, D = -4.
  3. Apply the distance formula: d=∣6(2)+(−3)(5)+2(−3)−4∣62+(−3)2+22d = \frac{|6(2) + (-3)(5) + 2(-3) - 4|}{\sqrt{6^2 + (-3)^2 + 2^2}}
  4. Calculate the numerator: ∣12−15−6−4∣=∣−13∣=13|12 - 15 - 6 - 4| = |-13| = 13.
  5. Calculate the denominator: 36+9+4=49=7\sqrt{36 + 9 + 4} = \sqrt{49} = 7.
  6. Therefore, d=137d = \frac{13}{7} units.

Explanation:

This solution uses the Cartesian distance formula. We substitute the point's coordinates into the plane's linear expression, take the absolute value to ensure a positive distance, and divide by the magnitude of the normal vector.

Problem 2:

Find the distance between the parallel planes 2x−y+2z+3=02x - y + 2z + 3 = 0 and 4x−2y+4z+15=04x - 2y + 4z + 15 = 0.

Solution:

  1. Express both equations with the same coefficients for x,y,zx, y, z. Divide the second equation by 22: 2x−y+2z+152=02x - y + 2z + \frac{15}{2} = 0.
  2. Now we have A=2,B=−1,C=2A=2, B=-1, C=2.
  3. Identify the constants: D1=3D_1 = 3 and D2=152=7.5D_2 = \frac{15}{2} = 7.5.
  4. Use the parallel plane distance formula: d=∣D1−D2∣A2+B2+C2d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}} d=∣3−7.5∣22+(−1)2+22d = \frac{|3 - 7.5|}{\sqrt{2^2 + (-1)^2 + 2^2}} d=∣−4.5∣4+1+4=4.53=1.5d = \frac{|-4.5|}{\sqrt{4 + 1 + 4}} = \frac{4.5}{3} = 1.5 units.

Explanation:

To find the distance between parallel planes, their x,y,zx, y, z coefficients must be identical first. Once normalized, the distance is the absolute difference between their constant terms divided by the length of the common normal vector.