Projection of Points and Lines on a Plane
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Direct answer
What does a plane do to a line that meets it? It squashes the line onto the surface: the projection of a line onto a plane is found by projecting any two of its points (feet of perpendiculars) and joining them, and the projected direction is b − (b·n̂)n̂. The scalar projection of a segment PQ on a second line with direction b is PQ·b̂ = (PQ·b)/|b| — one dot product, one division. A segment of length L making angle θ with a plane projects to length L cos θ. Degenerate cases are the exam's favourites: a line perpendicular to the plane projects to a single point, and a line already lying in the plane is its own projection.
What you must remember
- Projection of a vector a on b: the vector (a·b̂)b̂ with scalar part a·b̂ = (a·b)/|b| — dividing by |b| is what turns a dot product into a projection.
- Projection of a segment on a line: for P1, P2 and direction b, the scalar is ((x2 − x1)b1 + (y2 − y1)b2 + (z2 − z1)b3)/|b|; take the modulus for the length.
- On a plane by angle: length L at inclination θ to the plane projects to L cos θ — with the plane's surface, not its normal.
- Projected direction: b − (b·n̂)n̂ removes the normal component; if b ⊥ n this returns b unchanged (line parallel to plane).
- Point on plane: the foot of the perpendicular from P(x1, y1, z1) to ax + by + cz + d = 0 is P − s(a, b, c) with s = (ax1 + by1 + cz1 + d)/(a² + b² + c²).
- Degenerate cases: b parallel to n collapses the projection to a point; b in the plane leaves the line intact.
- Symmetric-form cue: the moment a question says "projection on the line (x − a)/l = ...", read b = (l, m, n) and compute one dot product.
Two projections worth mastering
Project the segment from P(1, 2, 3) to Q(2, −1, 4) onto the line with direction b = (2, 3, 6). The segment vector is PQ = (1, −3, 1). Its dot product with b is 1 × 2 + (−3) × 3 + 1 × 6 = 2 − 9 + 6 = −1, and |b| = √(4 + 9 + 36) = 7, so the scalar projection is −1/7 — a length of 1/7, with the negative sign recording that PQ leans against the direction of travel. That tiny value is the geometry talking: the segment is almost perpendicular to the line. Contrast the plane version: a rod of length 7 leaning at 30° to a floor casts a shadow of length 7 cos 30° = 7√3/2 ≈ 6.06 on the floor; tilt it to 60° and the shadow shrinks to 7/2. Same rod, different inclination, and the formula never needs the rod's actual endpoints — the angle carries all the information the plane question wants.
Small formula, frequent errors
JEE Main asks the scalar projection as a numerical answer, and the two standard losses are omitting the division by |b| (candidates report the raw dot product, here −1, as the projection) and forgetting that projecting onto a unit vector skips the division — the formula and the data must agree before computing. Options include both the signed value and its modulus; read which the question wants. JEE Advanced prefers the composite: find the projection of one line on another when both are given in symmetric form with different points, requiring two subtractions before the single dot product, or the projection of a segment onto a plane, where you need the angle between segment and plane rather than the normal — the sine-versus-cosine complement from the angle chapter resurfaces here as cos θ against the surface. The narrative discipline that protects marks: name the vector, name the target direction, divide by the target's length, and only then interpret the sign.
Frequently asked questions
What is the projection of vector a on vector b?
The vector (a·b/|b|²) b, whose length is |a·b|/|b| — the shadow a casts along b.
How long is the projection of a segment on a line?
|(P2 − P1)·b|/|b| for direction b: take the dot product of the segment with the direction and divide by the direction's length.
How does a length project onto a plane?
A segment of length L inclined at θ to the plane projects to L cos θ on the plane — the angle is with the surface itself.
What happens to a line perpendicular to the plane?
Its direction is entirely normal, so the projected direction b − (b·n̂)n̂ becomes zero and the line collapses to a single point, the foot.
How is the projection of an entire line on a plane constructed?
Project two points of the line onto the plane (feet of perpendiculars) and join them; the join is the projected line.