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Three-Dimensional Geometry - Equation of a Plane (Vector and Cartesian)

Grade 12ICSE

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

🔑Concepts

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A plane is defined as a flat surface such that if any two points are taken on it, the line segment joining them lies completely on the surface. Visually, it can be imagined as an infinite, perfectly flat sheet extending in all directions within three-dimensional space.

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The Normal Vector (n⃗\vec{n}) is a vector perpendicular to the plane. This is the most crucial identifying feature of a plane; you can visualize it as a flagpole standing perfectly vertical on a flat ground. Every vector lying on the plane is orthogonal to this normal vector.

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The Normal Form of a plane represents the plane in terms of its unit normal vector (n^\hat{n}) and its perpendicular distance (dd) from the origin. Visually, dd is the length of the shortest path from the origin (0,0,0)(0,0,0) to the plane's surface.

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A plane is uniquely determined if it passes through a specific point AA and is perpendicular to a given vector n⃗\vec{n}. If you pick any arbitrary point PP on the plane, the vector AP⃗\vec{AP} must be perpendicular to n⃗\vec{n}, satisfying the condition (r⃗−a⃗)⋅n⃗=0(\vec{r} - \vec{a}) \cdot \vec{n} = 0.

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Three non-collinear points A,B,A, B, and CC define a unique plane. This is similar to how a tripod stands stable on a floor; the three legs (points) define the flat surface. If the points were collinear (on a single line), infinitely many planes could rotate around that line like pages in a book.

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The Intercept Form of a plane is used when the plane intersects the coordinate axes. If it cuts the x,y,x, y, and zz axes at distances a,b,a, b, and cc from the origin respectively, it creates a visual 'corner cut' or a triangular facet across the axes.

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The Angle between two planes 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 planes are parallel if n1⃗\vec{n_1} is a scalar multiple of n2⃗\vec{n_2}, and they are perpendicular if their dot product n1⃗⋅n2⃗=0\vec{n_1} \cdot \vec{n_2} = 0.

📐Formulae

Vector equation of a plane in normal form: r⃗⋅n^=d\vec{r} \cdot \hat{n} = d

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 point a⃗\vec{a} and normal to n⃗\vec{n}: (r⃗−a⃗)⋅n⃗=0 or r⃗⋅n⃗=a⃗⋅n⃗(\vec{r} - \vec{a}) \cdot \vec{n} = 0 \text{ or } \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}

Cartesian equation of a plane passing through (x1,y1,z1)(x_1, y_1, z_1) with normal vector 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

General Cartesian equation of a plane: Ax+By+Cz+D=0Ax + By + Cz + D = 0

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

Equation of a plane passing through three non-collinear points (x1,y1,z1)(x_1, y_1, z_1), (x2,y2,z2)(x_2, y_2, z_2), and (x3,y3,z3)(x_3, y_3, z_3): ∣x−x1y−y1z−z1x2−x1y2−y1z2−z1x3−x1y3−y1z3−z1∣=0\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0

Perpendicular distance of a point (x1,y1,z1)(x_1, y_1, z_1) from the plane Ax+By+Cz+D=0Ax + By + Cz + D = 0: p=∣Ax1+By1+Cz1+D∣A2+B2+C2p = \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 position vector of the given point be 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 of the line: n⃗=2i^+3j^−k^\vec{n} = 2\hat{i} + 3\hat{j} - \hat{k}.
  3. The vector equation is (r⃗−a⃗)⋅n⃗=0⇒r⃗⋅n⃗=a⃗⋅n⃗(\vec{r} - \vec{a}) \cdot \vec{n} = 0 \Rightarrow \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. Cartesian Equation: Substitute r⃗=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k} to get 2x+3y−z=202x + 3y - z = 20.

Explanation:

This problem uses the point-normal form. We identify the given point as a⃗\vec{a} and the perpendicular direction as the normal vector n⃗\vec{n}. The dot product of the arbitrary position vector and the normal equals the dot product of the known point and the normal.

Problem 2:

Find the equation of the plane that makes intercepts 2,3,2, 3, and 44 on the x,y,x, y, and zz axes respectively.

Solution:

  1. The intercepts are given as a=2,b=3,c=4a = 2, b = 3, c = 4.
  2. Use the intercept form: xa+yb+zc=1\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1.
  3. Substitute the values: x2+y3+z4=1\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1.
  4. To clear the fractions, find the LCM of 2,3,2, 3, and 44, which is 1212.
  5. Multiply the entire equation by 1212: 6x+4y+3z=126x + 4y + 3z = 12.

Explanation:

The intercept form is the most efficient way to find the equation when the points where the plane crosses the axes are known. We simply substitute the intercept values into the standard xa+yb+zc=1\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 formula and simplify to general form.