Vector Equations of Lines and Planes
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Direct answer
Points, lines and planes in three dimensions speak most fluently in vectors: a line through point a with direction b is r = a + λb for real λ; a line through two points a and b is r = a + λ(b − a); a plane with normal n through point a satisfies (r − a)·n = 0, written r·n = a·n. The plane through three non-collinear points a, b, c uses the normal (b − a) × (c − a), giving r·[(b − a) × (c − a)] = a·[(b − a) × (c − a)]. Angles follow from dot products: between two lines cos θ = |b₁·b₂|/(|b₁||b₂|), between line and plane sin θ = |b·n|/(|b||n|), between planes cos θ = |n₁·n₂|/(|n₁||n₂|) — the sine in the line-plane case is the detail exams harvest.
What you must remember
- Line forms: r = a + λb (point-direction); r = a + λ(b − a) (two points); Cartesian (x − x₁)/a = (y − y₁)/b = (z − z₁)/c with direction ratios (a, b, c).
- Plane forms: r·n̂ = d (unit normal, d = distance from origin); r·n = d (any normal); intercept form x/a + y/b + z/c = 1 with intercepts a, b, c.
- Three-point plane: normal = (b − a) × (c − a); a plane through the line of intersection of two planes is S₁ + λS₂ = 0 (the family-of-planes trick).
- Angle formulas: line-line uses cos θ = |b₁·b₂|/(|b₁||b₂|); line-plane uses sin θ = |b·n|/(|b||n|); plane-plane uses cos θ = |n₁·n₂|/(|n₁||n₂|).
- Perpendicularity/parallelism translations: lines parallel iff b₁ ∥ b₂; line perpendicular to plane iff b ∥ n; planes parallel iff n₁ ∥ n₂; a line lies in a plane iff a satisfies the plane and b·n = 0.
- Coplanarity test for two lines: the lines r = a₁ + λb₁ and r = a₂ + μb₂ are coplanar iff (a₂ − a₁)·(b₁ × b₂) = 0, the scalar triple product condition.
Building the forms
Construct the plane through A(1, 1, 0), B(2, 0, 1), C(3, 1, 1). Two in-plane vectors: AB = (1, −1, 1) and AC = (2, 0, 1). Their cross product: i-component (−1·1 − 1·0) = −1, j-component −(1·1 − 1·2) = 1, k-component (1·0 − (−1)·2) = 2, so n = (−1, 1, 2). The plane is −(x − 1) + (y − 1) + 2z = 0, i.e. −x + y + 2z = 0; the unused point C confirms it, since −3 + 1 + 2 = 0. One construction, whole toolkit: vectors from one vertex, cross product normal, point form, verification.
The family trick completes the picture: all planes through the line x + y + z = 1, 2x − y + 3z = 5 are (x + y + z − 1) + λ(2x − y + 3z − 5) = 0. To pick the member through (1, 0, 2): 1 + 0 + 2 − 1 + λ(2 − 0 + 6 − 5) = 2 + 3λ = 0, λ = −2/3, and expanding gives the specific plane. JEE asks exactly this — "the plane through the line of intersection and the point" — and the family method finishes it in three lines.
Formula choice traps
The recurring error is the line-plane angle: students reflexively use cosine with the normal, landing on the complement — the sine formula exists because the line meets the plane, not the normal. The second trap is normalising too early: r·n̂ = d needs the unit normal — using an unnormalised n changes d — so either normalise first or use r·n = a·n. Third, converting vector to Cartesian form: from r = (2i − j + k) + λ(i + j − 2k), write x = 2 + λ, y = −1 + λ, z = 1 − 2λ and eliminate λ pairwise. Main-level questions convert forms and compute one angle; Advanced compose coplanarity with distances, where the scalar triple product is the entry ticket.
Frequently asked questions
What is the vector equation of a line through two given points?
r = a + λ(b − a), where a and b are the position vectors of the points; λ runs over all reals.
How do you find the plane through three points?
Form vectors from one point to the other two, take their cross product as the normal n, then write (r − a)·n = 0 — checking first that the points are not collinear.
Why does the line-plane angle use sine?
Because the angle between the line and the plane is the complement of the angle between the line's direction and the plane's normal, so sin θ = |b·n|/(|b||n|).
What is the family of planes through a line of intersection?
S₁ + λS₂ = 0, where S₁ = 0 and S₂ = 0 are two planes through the line; λ selects the member satisfying one extra condition.
How do you test whether two lines in space are coplanar?
Check the scalar triple product (a₂ − a₁)·(b₁ × b₂) = 0; coplanar lines then either intersect or are parallel.