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Concept 05 / 06Three Dimensional Geometry

Equation of a Plane: Normal, Intercept, Three-Point Forms and Angles Between Planes

The plane through a point with a given normal, the general equation ax + by + cz + d = 0, normal form r.n = p, intercept form, the plane through three points, the angle between two planes, parallel and perpendicular planes, and the family of planes through a line.

Figure 1
A plane is determined by a point on it and a normal vector (a vector perpendicular to every line in the plane). Point-normal form. The plane through a⃗\vec{a} with normal n⃗\vec{n} consists of the points r⃗\vec{r} for which r⃗−a⃗\vec{r} - \vec{a} lies in the plane, i.e. is perpendicular to n⃗\vec{n}: (r⃗−a⃗)⋅n⃗=0(\vec{r} - \vec{a}) \cdot \vec{n} = 0. In Cartesian form, with n⃗=(a,b,c)\vec{n} = (a, b, c) and the point (x1,y1,z1)(x_{1}, y_{1}, z_{1}): a(x−x1)+b(y−y1)+c(z−z1)=0a(x - x_{1}) + b(y - y_{1}) + c(z - z_{1}) = 0. General form: ax+by+cz+d=0ax + by + cz + d = 0. The coefficients of x,y,zx, y, z are the direction ratios of the normal. This is the most important reading skill in the topic. Normal form: r⃗⋅n^=p\vec{r} \cdot \hat{n} = p, i.e. lx+my+nz=plx + my + nz = p with (l,m,n)(l, m, n) the direction cosines of the normal and p≥0p \ge 0 the perpendicular distance from the origin. From the general form, divide by ±a2+b2+c2\pm\sqrt{a^{2} + b^{2} + c^{2}}, choosing the sign that makes the right side positive. Intercept form: xa+yb+zc=1\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1, cutting the axes at (a,0,0)(a, 0, 0), (0,b,0)(0, b, 0) and (0,0,c)(0, 0, c), as in the figure. The area of the triangle cut off in the first octant and the volume of the tetrahedron with the origin (∣abc∣6\frac{|abc|}{6}) are common questions. Plane through three points A,B,CA, B, C. The normal is AB⃗×AC⃗\vec{AB} \times \vec{AC}. Equivalently, [r⃗−a⃗  b⃗−a⃗  c⃗−a⃗]=0[\vec{r} - \vec{a}\ \ \vec{b} - \vec{a}\ \ \vec{c} - \vec{a}] = 0, i.e. ∣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. Four points are coplanar iff the fourth satisfies this equation. Special planes. The xyxy-plane is z=0z = 0. A plane parallel to the zz-axis has no zz term: ax+by+d=0ax + by + d = 0. A plane through the origin has d=0d = 0. Angle between two planes is the angle between their normals: cos⁡θ=∣n⃗1⋅n⃗2∣∣n⃗1∣∣n⃗2∣=∣a1a2+b1b2+c1c2∣∑a12∑a22\cos\theta = \frac{|\vec{n}_{1} \cdot \vec{n}_{2}|}{|\vec{n}_{1}||\vec{n}_{2}|} = \frac{|a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2}|}{\sqrt{\sum a_{1}^{2}}\sqrt{\sum a_{2}^{2}}}. Perpendicular planes: a1a2+b1b2+c1c2=0a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0. Parallel planes: a1a2=b1b2=c1c2\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}. A plane parallel to ax+by+cz+d=0ax + by + cz + d = 0 is ax+by+cz+k=0ax + by + cz + k = 0. Family of planes through a line. If a line is the intersection of P1≡a1x+b1y+c1z+d1=0P_{1} \equiv a_{1}x + b_{1}y + c_{1}z + d_{1} = 0 and P2=0P_{2} = 0, then every plane through that line (except P2P_{2} itself) is P1+λP2=0P_{1} + \lambda P_{2} = 0. Impose one more condition (passing through a point, or perpendicular to a given plane) to find λ\lambda. This avoids computing the line explicitly. Plane through a point containing a given line, or containing two intersecting lines: the normal is the cross product of two direction vectors lying in the plane. Key Takeaways: - (r⃗−a⃗)⋅n⃗=0(\vec{r} - \vec{a}) \cdot \vec{n} = 0; a(x−x1)+b(y−y1)+c(z−z1)=0a(x - x_{1}) + b(y - y_{1}) + c(z - z_{1}) = 0. - In ax+by+cz+d=0ax + by + cz + d = 0, (a,b,c)(a, b, c) is the normal. - Normal form lx+my+nz=plx + my + nz = p; intercept form xa+yb+zc=1\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1. - Three points: normal AB⃗×AC⃗\vec{AB} \times \vec{AC}, or the determinant equation. - Angle between planes == angle between normals; perpendicular iff ∑a1a2=0\sum a_{1}a_{2} = 0. - Planes through the line P1=P2=0P_{1} = P_{2} = 0: P1+λP2=0P_{1} + \lambda P_{2} = 0. - Pitfall: treating a direction vector lying in the plane as its normal. Brain-Teaser 1: Find the plane through (1,1,1)(1, 1, 1), (1,−1,2)(1, -1, 2) and (−2,−2,2)(-2, -2, 2). Solution: AB⃗=(0,−2,1)\vec{AB} = (0, -2, 1), AC⃗=(−3,−3,1)\vec{AC} = (-3, -3, 1). Normal: AB⃗×AC⃗=((−2)(1)−(1)(−3),(1)(−3)−(0)(1),(0)(−3)−(−2)(−3))=(1,−3,−6)\vec{AB} \times \vec{AC} = ((-2)(1) - (1)(-3), (1)(-3) - (0)(1), (0)(-3) - (-2)(-3)) = (1, -3, -6). Plane: (x−1)−3(y−1)−6(z−1)=0(x - 1) - 3(y - 1) - 6(z - 1) = 0, i.e. x−3y−6z+8=0x - 3y - 6z + 8 = 0. Check with CC: −2+6−12+8=0-2 + 6 - 12 + 8 = 0. Correct. Brain-Teaser 2: Find the plane through the line of intersection of x+y+z=6x + y + z = 6 and 2x+3y+4z+5=02x + 3y + 4z + 5 = 0 that passes through (1,1,1)(1, 1, 1). Solution: The family: (x+y+z−6)+λ(2x+3y+4z+5)=0(x + y + z - 6) + \lambda(2x + 3y + 4z + 5) = 0. At (1,1,1)(1, 1, 1): −3+14λ=0-3 + 14\lambda = 0, so λ=314\lambda = \frac{3}{14}. Multiplying by 1414: 14(x+y+z−6)+3(2x+3y+4z+5)=014(x + y + z - 6) + 3(2x + 3y + 4z + 5) = 0, i.e. 20x+23y+26z−69=020x + 23y + 26z - 69 = 0. Check: 20+23+26=6920 + 23 + 26 = 69. The family method needed no explicit direction for the line. Brain-Teaser 3: A plane meets the axes at A,B,CA, B, C so that the centroid of triangle ABCABC is (1,2,3)(1, 2, 3). Find the plane and the volume of the tetrahedron OABCOABC. Solution: With intercepts a,b,ca, b, c, the centroid is (a3,b3,c3)=(1,2,3)(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}) = (1, 2, 3), so a=3a = 3, b=6b = 6, c=9c = 9. The plane is x3+y6+z9=1\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1, i.e. 6x+3y+2z=186x + 3y + 2z = 18. The volume is abc6=1626=27\frac{abc}{6} = \frac{162}{6} = 27. The intercept form turns the centroid condition into three one-line equations.