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Concept 06 / 06Three Dimensional Geometry
Coplanarity of Lines, Planes Containing Lines and the Sphere
The condition for two lines to be coplanar and the plane containing them, the plane through a line and a point or parallel to another line, the equation of a sphere, its centre and radius, tangent planes, and the circle cut from a sphere by a plane.
Figure 1
Coplanar lines. Two lines r=a1+λb1 and r=a2+μb2 lie in one plane iff
[a2−a1b1b2]=0, i.e. x2−x1a1a2y2−y1b1b2z2−z1c1c2=0.
If they are not parallel, the plane containing them passes through a1 with normal b1×b2:
x−x1a1a2y−y1b1b2z−z1c1c2=0.
If they are parallel (and distinct), the plane contains a1, the direction b and the vector a2−a1, so its normal is b×(a2−a1).
Other planes built from lines.
Plane containing a line and a point Q not on it: normal b×(q−a).
Plane containing one line and parallel to another: normal b1×b2, through a point of the first line. The distance between two skew lines is the distance from any point of the second line to this plane, which is another way to derive the shortest-distance formula.
Plane containing a line and perpendicular to a plane with normal n: normal b×n.
The sphere. A sphere is the set of points at a fixed distance R (the radius) from a fixed point C (the centre): ∣r−c∣=R. In Cartesian form,
(x−α)2+(y−β)2+(z−γ)2=R2,
or in general form x2+y2+z2+2ux+2vy+2wz+d=0, with centre (−u,−v,−w) and radius u2+v2+w2−d. A second-degree equation represents a sphere iff the coefficients of x2,y2,z2 are equal and there are no xy, yz or zx terms. The sphere on the segment joining A and B as diameter is (r−a)⋅(r−b)=0, the same 'angle in a semicircle' argument as for the circle.
Plane sections and tangent planes. A plane at distance d from the centre cuts the sphere in a circle of radius R2−d2 if d<R, as the figure shows; its centre is the foot of the perpendicular from the sphere's centre to the plane. The plane touches the sphere if d=R, and misses it if d>R. The tangent plane at a point p of the sphere is perpendicular to the radius: (r−p)⋅(p−c)=0. In Cartesian form it is obtained by the same replacement rule as for the circle: xx1+yy1+zz1+u(x+x1)+v(y+y1)+w(z+z1)+d=0.
Key Takeaways:
- Coplanar lines: [a2−a1b1b2]=0; plane normal b1×b2.
- Plane through a line and a point: normal b×(q−a).
- Plane through line 1 parallel to line 2: normal b1×b2.
- Sphere: centre (−u,−v,−w), radius u2+v2+w2−d.
- Diametric sphere: (r−a)⋅(r−b)=0.
- Plane section radius R2−d2; tangent plane iff d=R.
- Pitfall: using b1×b2 as the normal when the two lines are parallel (it is 0).
Brain-Teaser 1: Show that the lines −3x+3=1y−1=5z−5 and −1x+1=2y−2=5z−5 are coplanar, and find the plane containing them.
Solution: a2−a1=(−1+3,2−1,5−5)=(2,1,0). The determinant 2−3−1112055=2(5−10)−1(−15+5)+0=−10+10=0, so they are coplanar. The normal is (−3,1,5)×(−1,2,5)=(5−10,−5+15,−6+1)=(−5,10,−5), or (1,−2,1). Through (−3,1,5): (x+3)−2(y−1)+(z−5)=0, i.e. x−2y+z=0. Both base points satisfy it: −3−2+5=0 and −1−4+5=0.
Brain-Teaser 2: Find the radius of the circle in which the plane x+2y+2z=15 cuts the sphere x2+y2+z2−2y−4z=11.
Solution: The sphere has centre (0,1,2) and radius 1+4+11=4. The distance from the centre to the plane is 3∣0+2+4−15∣=3. The circle's radius is 16−9=7. Its centre is the foot of the perpendicular from (0,1,2): moving 3 units along the unit normal 31(1,2,2) gives (1,3,4), which lies on the plane.
Brain-Teaser 3: Find the equation of the plane that contains the line 2x−1=−1y+1=4z−3 and is perpendicular to the plane x+2y+z=12.
Solution: The required normal is perpendicular to both the line's direction (2,−1,4) and the given normal (1,2,1): (2,−1,4)×(1,2,1)=(−1−8,4−2,4+1)=(−9,2,5). Through (1,−1,3): −9(x−1)+2(y+1)+5(z−3)=0, i.e. 9x−2y−5z+4=0. The projection of the line on the given plane is the intersection of these two planes, which is how projection questions are usually solved.
