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Three Dimensional Geometry - Distance of a point from a plane

Grade 12CBSE

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

🔑Concepts

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The distance of a point from a plane is defined as the length of the perpendicular segment dropped from the point to the plane. Geometrically, if you imagine a point PP suspended in space and a flat surface representing the plane, the distance is the shortest straight-line path from PP to the surface, meeting it at a 90∘90^{\circ} angle.

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In the Vector representation, the plane is often given by the equation r⃗⋅n⃗=d\vec{r} \cdot \vec{n} = d, where n⃗\vec{n} is the normal vector perpendicular to the plane. The distance is found by projecting the position vector of the point onto this normal vector.

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In Cartesian form, the plane is expressed as Ax+By+Cz+D=0Ax + By + Cz + D = 0. Here, the coefficients (A,B,C)(A, B, C) represent the direction ratios of the normal to the plane, indicating the direction that is perfectly 'upright' relative to the plane's surface.

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The 'Foot of the Perpendicular' is the specific point MM on the plane where the perpendicular line from point PP intersects the plane. Visualizing this, MM is the shadow of point PP if a light source were placed infinitely far away directly above PP perpendicular to the plane.

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The distance is always a non-negative scalar quantity. When using the formula, we apply absolute value signs to the numerator to ensure that even if the calculation yields a negative value (indicating the point is on the opposite side of the normal's direction), the distance remains positive.

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Distance from the Origin is a special case of the general formula. Since the coordinates of the origin are (0,0,0)(0, 0, 0), the variables in the Cartesian equation vanish, leaving the distance as the magnitude of the constant term divided by the magnitude of the normal vector.

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The relationship between the point and the plane can be visualized using the normal vector. If the distance is zero, it implies the point PP lies exactly on the plane, satisfying the plane's equation.

📐Formulae

Distance of a point with position vector a⃗\vec{a} from the plane r⃗⋅n⃗=d\vec{r} \cdot \vec{n} = d: p=∣a⃗⋅n⃗−d∣∣n⃗∣p = \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|}

Distance of a point P(x1,y1,z1)P(x_1, y_1, z_1) from the plane 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}}

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

Magnitude of the normal vector n⃗=Ai^+Bj^+Ck^\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}: ∣n⃗∣=A2+B2+C2|\vec{n}| = \sqrt{A^2 + B^2 + C^2}

💡Examples

Problem 1:

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

Solution:

  1. Identify the coordinates of the point: (x1,y1,z1)=(2,5,−3)(x_1, y_1, z_1) = (2, 5, -3). \n2. Identify the coefficients from the plane equation Ax+By+Cz+D=0Ax + By + Cz + D = 0: A=6,B=−3,C=2,D=−4A = 6, B = -3, C = 2, D = -4. \n3. Use the Cartesian distance formula: d=∣Ax1+By1+Cz1+D∣A2+B2+C2d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}. \n4. Substitute the values: 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}}. \n5. Simplify the numerator: ∣12−15−6−4∣=∣−13∣=13|12 - 15 - 6 - 4| = |-13| = 13. \n6. Simplify the denominator: 36+9+4=49=7\sqrt{36 + 9 + 4} = \sqrt{49} = 7. \n7. Final distance d=137d = \frac{13}{7} units.

Explanation:

This solution applies the Cartesian distance formula by substituting the point's coordinates into the plane's linear expression and dividing by the magnitude of the normal vector (6,−3,2)(6, -3, 2).

Problem 2:

Find the distance of a point with position vector a⃗=2i^+j^−k^\vec{a} = 2\hat{i} + \hat{j} - \hat{k} from the plane r⃗⋅(3i^−4j^+12k^)=9\vec{r} \cdot (3\hat{i} - 4\hat{j} + 12\hat{k}) = 9.

Solution:

  1. Identify a⃗=2i^+j^−k^\vec{a} = 2\hat{i} + \hat{j} - \hat{k}, n⃗=3i^−4j^+12k^\vec{n} = 3\hat{i} - 4\hat{j} + 12\hat{k}, and d=9d = 9. \n2. Calculate the dot product a⃗⋅n⃗\vec{a} \cdot \vec{n}: (2)(3)+(1)(−4)+(−1)(12)=6−4−12=−10(2)(3) + (1)(-4) + (-1)(12) = 6 - 4 - 12 = -10. \n3. Calculate the magnitude of the normal vector ∣n⃗∣|\vec{n}|: 32+(−4)2+122=9+16+144=169=13\sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13. \n4. Use the vector distance formula: p=∣a⃗⋅n⃗−d∣∣n⃗∣p = \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|}. \n5. Substitute the values: p=∣−10−9∣13=∣−19∣13=1913p = \frac{|-10 - 9|}{13} = \frac{|-19|}{13} = \frac{19}{13} units.

Explanation:

This approach uses the vector form of the distance formula. We find the scalar projection of the point's position vector onto the normal direction and adjust for the plane's offset from the origin.