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Three Dimensional Geometry - Cartesian and vector equation of a plane

Grade 12CBSE

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

🔑Concepts

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Normal Form of a Plane: A plane is uniquely determined if its perpendicular distance from the origin (dd) and the direction cosines (l,m,nl, m, n) of the normal to the plane are known. Visually, imagine a flat sheet in 3D space with a perpendicular stick (the normal) pointing from the origin (0,0,0)(0,0,0) to the plane surface with length dd.

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Point-Normal Equation: A plane can be defined by a single point AA with position vector a⃗\vec{a} through which it passes, and a vector n⃗\vec{n} that is perpendicular (normal) to the plane's surface. Geometrically, any vector lying on the plane starting from point AA will be at a 90∘90^{\circ} angle to the normal vector n⃗\vec{n}.

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Plane through Three Non-Collinear Points: Just as a line is determined by two points, a unique plane is determined by three points that do not lie on the same straight line. If you visualize three points in space, they form a triangle that defines the 'flatness' and orientation of the entire plane extending infinitely in all directions.

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Intercept Form: This concept describes a plane by where it crosses the X,Y,X, Y, and ZZ axes. The distances from the origin to these intersection points are called intercepts (a,b,ca, b, c). Visually, the plane cuts through the corner of a room, creating a triangular face between the three coordinate axes.

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Angle Between Two Planes: The tilt between two flat surfaces is defined as the angle between their respective normal vectors. If the normal vectors are n1⃗\vec{n_1} and n2⃗\vec{n_2}, the angle θ\theta is found using the dot product. If the planes are perpendicular, their normals are perpendicular; if the planes are parallel, their normals are parallel.

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Distance of a Point from a Plane: This represents the shortest distance from a specific point P(x1,y1,z1)P(x_1, y_1, z_1) to the surface of the plane. Visually, this is the length of the perpendicular line segment dropped from the point onto the plane surface.

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Coplanarity of Two Lines: Two lines in 3D space are coplanar if they lie in the exact same plane. For this to happen, either the lines must intersect at a point or they must be parallel. If they are 'skew lines' (neither parallel nor intersecting), they cannot share a single plane.

📐Formulae

Vector equation of a plane in normal form: r⃗⋅n^=d\vec{r} \cdot \hat{n} = d, where n^\hat{n} is the unit normal vector.

Cartesian equation of a plane in normal form: lx+my+nz=dlx + my + nz = d, where l,m,nl, m, n are direction cosines.

Vector equation of a plane passing through a⃗\vec{a} and perpendicular to n⃗\vec{n}: (r⃗−a⃗)⋅n⃗=0(\vec{r} - \vec{a}) \cdot \vec{n} = 0

Cartesian equation of a plane through (x1,y1,z1)(x_1, y_1, z_1) with normal direction ratios (A,B,C)(A, B, C): A(x−x1)+B(y−y1)+C(z−z1)=0A(x - x_1) + B(y - y_1) + C(z - z_1) = 0

Intercept form: xa+yb+zc=1\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1

Angle θ\theta between two planes r⃗⋅n1⃗=d1\vec{r} \cdot \vec{n_1} = d_1 and r⃗⋅n2⃗=d2\vec{r} \cdot \vec{n_2} = d_2: cos⁡θ=∣n1⃗⋅n2⃗∣∣n1⃗∣∣n2⃗∣\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}

Distance of point (x1,y1,z1)(x_1, y_1, z_1) from 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}}

💡Examples

Problem 1:

Find the vector and Cartesian equations of the plane which passes through the point (5,2,−4)(5, 2, -4) and is perpendicular to the line with direction ratios (2,3,−1)(2, 3, -1).

Solution:

  1. Let the given point be A(5,2,−4)A(5, 2, -4), so its position vector is a⃗=5i^+2j^−4k^\vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k}.
  2. The normal vector n⃗\vec{n} is given by the direction ratios: n⃗=2i^+3j^−k^\vec{n} = 2\hat{i} + 3\hat{j} - \hat{k}.
  3. The vector equation is (r⃗−a⃗)⋅n⃗=0(\vec{r} - \vec{a}) \cdot \vec{n} = 0, which simplifies to r⃗⋅n⃗=a⃗⋅n⃗\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}.
  4. Calculate a⃗⋅n⃗=(5)(2)+(2)(3)+(−4)(−1)=10+6+4=20\vec{a} \cdot \vec{n} = (5)(2) + (2)(3) + (-4)(-1) = 10 + 6 + 4 = 20.
  5. Vector Equation: r⃗⋅(2i^+3j^−k^)=20\vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 20.
  6. For Cartesian equation, substitute r⃗=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}: 2x+3y−z=202x + 3y - z = 20.

Explanation:

We use the point-normal form of the plane equation. The direction ratios of the perpendicular line serve as the components of the normal vector n⃗\vec{n}.

Problem 2:

Find the distance of the point (2,5,−3)(2, 5, -3) from the plane r⃗⋅(6i^−3j^+2k^)=4\vec{r} \cdot (6\hat{i} - 3\hat{j} + 2\hat{k}) = 4.

Solution:

  1. Convert the plane equation to Cartesian form: 6x−3y+2z−4=06x - 3y + 2z - 4 = 0.
  2. Identify coordinates of the point: (x1,y1,z1)=(2,5,−3)(x_1, y_1, z_1) = (2, 5, -3).
  3. Identify plane coefficients: A=6,B=−3,C=2,D=−4A = 6, B = -3, C = 2, D = -4.
  4. 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}}.
  5. d=∣12−15−6−4∣36+9+4=∣−13∣49=137d = \frac{|12 - 15 - 6 - 4|}{\sqrt{36 + 9 + 4}} = \frac{|-13|}{\sqrt{49}} = \frac{13}{7}.
  6. The distance is 137\frac{13}{7} units.

Explanation:

The distance is calculated by substituting the point coordinates into the general Cartesian form of the plane and dividing by the magnitude of the normal vector.