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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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The Normal Form of a plane is defined by its distance dd from the origin and a unit normal vector n^\hat{n} perpendicular to the plane. The vector equation is r⃗⋅n^=d\vec{r} \cdot \hat{n} = d. In Cartesian form, this is lx+my+nz=dlx + my + nz = d, where l,m,nl, m, n are the direction cosines of the normal.

A plane in 3D space showing a normal vector originating from the origin.
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A plane is uniquely determined if it passes through a fixed point AA with position vector a⃗\vec{a} and is perpendicular to a given vector n⃗\vec{n}. Every point P(r⃗)P(\vec{r}) on the plane must satisfy the condition that the vector AP⃗\vec{AP} is perpendicular to n⃗\vec{n}, leading to (r⃗−a⃗)⋅n⃗=0(\vec{r} - \vec{a}) \cdot \vec{n} = 0.

Geometric representation of a plane passing through point A with normal vector n.
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The Intercept Form of the equation of a plane is xa+yb+zc=1\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1, where a,b,ca, b, c are the lengths of the intercepts made by the plane on the x,y,x, y, and zz axes respectively.

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The shortest distance from a point (x1,y1,z1)(x_1, y_1, z_1) to a plane Ax+By+Cz+D=0Ax + By + Cz + D = 0 is the length of the perpendicular dropped from the point to the plane.

📐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.

Problem 3:

Find the equation of the plane passing through the points P(1,1,−1)P(1, 1, -1), Q(6,4,−5)Q(6, 4, -5), and R(−4,−2,3)R(-4, -2, 3).

Three points P, Q, and R lying on a single straight line in space.

Solution:

The general equation of a plane passing through (x1,y1,z1)(x_1, y_1, z_1) is A(x−x1)+B(y−y1)+C(z−z1)=0A(x - x_1) + B(y - y_1) + C(z - z_1) = 0. Substituting P(1,1,−1)P(1, 1, -1): A(x−1)+B(y−1)+C(z+1)=0A(x - 1) + B(y - 1) + C(z + 1) = 0 ...(i) Since QQ and RR lie on the plane: A(6−1)+B(4−1)+C(−5+1)=0⇒5A+3B−4C=0A(6-1) + B(4-1) + C(-5+1) = 0 \Rightarrow 5A + 3B - 4C = 0 A(−4−1)+B(−2−1)+C(3+1)=0⇒−5A−3B+4C=0A(-4-1) + B(-2-1) + C(3+1) = 0 \Rightarrow -5A - 3B + 4C = 0 Notice these equations are dependent (one is the negative of the other). This suggests the points are collinear. However, testing the ratio: 5−5=3−3=−44\frac{5}{-5} = \frac{3}{-3} = \frac{-4}{4}. Since the points are collinear, infinitely many planes can pass through them. Usually, for a unique plane, we use the determinant: ∣x−1y−1z+16−14−1−5+1−4−1−2−13+1∣=0\begin{vmatrix} x - 1 & y - 1 & z + 1 \\ 6 - 1 & 4 - 1 & -5 + 1 \\ -4 - 1 & -2 - 1 & 3 + 1 \end{vmatrix} = 0 ∣x−1y−1z+153−4−5−34∣=0\begin{vmatrix} x - 1 & y - 1 & z + 1 \\ 5 & 3 & -4 \\ -5 & -3 & 4 \end{vmatrix} = 0 Since Row 2 and Row 3 are proportional, the determinant is 00 for any x,y,zx, y, z satisfying a specific linear relation, or in this case, it confirms the points do not form a unique plane.

Explanation:

To find a unique plane through three points, they must be non-collinear. If the determinant results in 0=00 = 0, the points are collinear.

Problem 4:

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.

Diagram showing the perpendicular distance from a point to a plane.

Solution:

The Cartesian equation of the plane is 6x−3y+2z−4=06x - 3y + 2z - 4 = 0. Given point (x1,y1,z1)=(2,5,−3)(x_1, y_1, z_1) = (2, 5, -3). A=6,B=−3,C=2,D=−4A = 6, B = -3, C = 2, D = -4. Distance p=∣6(2)−3(5)+2(−3)−4∣62+(−3)2+22p = \frac{|6(2) - 3(5) + 2(-3) - 4|}{\sqrt{6^2 + (-3)^2 + 2^2}} p=∣12−15−6−4∣36+9+4p = \frac{|12 - 15 - 6 - 4|}{\sqrt{36 + 9 + 4}} p=∣−13∣49=137p = \frac{|-13|}{\sqrt{49}} = \frac{13}{7} units.

Explanation:

Convert the vector equation to Cartesian form, then apply the perpendicular distance formula.