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Concept 02 / 06Three Dimensional Geometry
Equation of a Line in Space: Vector and Cartesian Forms
The line through a point parallel to a vector, the line through two points, the symmetric Cartesian form and general point, converting between forms, the angle between two lines, and finding the foot of the perpendicular and image of a point in a line.
Figure 1
A line in space is determined by a point on it and a direction.
Vector form. The line through the point with position vector a, parallel to b, is
r=a+λb, λ∈R.
Each value of λ gives one point, as in the figure. The line through two points a and c is r=a+λ(c−a).
Cartesian (symmetric) form. With a=(x1,y1,z1) and direction ratios b=(a,b,c), eliminating λ gives
ax−x1=by−y1=cz−z1.
If one direction ratio is 0, say c=0, the line lies in the plane z=z1; the symmetric form is still written with the 0 denominator, interpreted as z=z1. The line through two points: x2−x1x−x1=y2−y1y−y1=z2−z1z−z1.
The general point. Setting the common ratio equal to λ, every point of the line is (x1+aλ,y1+bλ,z1+cλ). This parametrisation is the key to almost every problem: to find where a line meets a plane, or the foot of a perpendicular, or the intersection of two lines, write the general point and impose the condition to find λ.
Reading the standard form carefully. The coefficients of x,y,z in the numerators must be 1. For example, 32x−1=4y+2=21−z must be rewritten as 23x−21=4y+2=−2z−1 before reading off the point (21,−2,1) and direction ratios (23,4,−2), or (3,8,−4).
Line as the intersection of two planes. A line given by two plane equations a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0 has direction along n1×n2 (perpendicular to both normals). A point on it is found by setting one coordinate (say z=0) and solving the two equations for the others.
Angle between two lines. With direction vectors b1,b2: cosθ=∣b1∣∣b2∣∣b1⋅b2∣.
Foot of the perpendicular from a point P to a line. Take the general point Q=a+λb of the line, and impose PQ⋅b=0. This gives λ, hence the foot Q, the perpendicular distance PQ, and the image of P in the line, 2Q−P. A formula for the distance: ∣b∣∣(p−a)×b∣ (area of the parallelogram divided by its base).
Key Takeaways:
- r=a+λb; through two points: r=a+λ(c−a).
- ax−x1=by−y1=cz−z1; general point (x1+aλ,y1+bλ,z1+cλ).
- Make the coefficients of x,y,z equal to 1 before reading off data.
- Intersection of two planes: direction n1×n2.
- Foot of perpendicular: general point Q with PQ⋅b=0; image 2Q−P.
- Distance from a point to a line: ∣b∣∣(p−a)×b∣.
- Pitfall: reading (1−z) as having direction ratio +1 instead of −1.
Brain-Teaser 1: Find the foot of the perpendicular from P(1,6,3) to the line 1x=2y−1=3z−2, and the image of P.
Solution: The general point is Q(λ,1+2λ,2+3λ). Then PQ=(λ−1,2λ−5,3λ−1), and PQ⋅(1,2,3)=λ−1+4λ−10+9λ−3=14λ−14=0, so λ=1 and Q=(1,3,5). The perpendicular distance is 0+9+4=13. The image is 2Q−P=(1,0,7).
Brain-Teaser 2: Find the line of intersection of the planes x+y+z=1 and 2x−y+z=3 in symmetric form.
Solution: Direction: n1×n2=(1,1,1)×(2,−1,1)=(1⋅1−1⋅(−1),1⋅2−1⋅1,1⋅(−1)−1⋅2)=(2,1,−3). A point: put z=0: x+y=1, 2x−y=3, so x=34, y=−31. The line is 2x−4/3=1y+1/3=−3z. Check: any point of it, for example λ=1 giving (310,32,−3), satisfies both plane equations.
Brain-Teaser 3: Find the angle between the lines 2x−1=−1y=1z+1 and 22x−1=13−y=2z.
Solution: The first has direction (2,−1,1). Rewrite the second: 1x−1/2=−1y−3=2z, with direction (1,−1,2). Then cosθ=66∣2+1+2∣=65, so θ=cos−165. Normalising the second equation was essential; using (2,1,2) from the unprocessed denominators would give a wrong angle.
