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MHT-CET Maths · Line and Plane

Foot of Perpendicular, Image, and Projection

One engine drives this whole subtopic: drop a perpendicular from a point to a line or a plane, locate its FOOT, and then either report the foot, double it across to get the mirror image (2F − P), or use a dot product to read off a projection length.

All Line and Plane notesMHT-CET Maths strategy

Why this matters

Every MHT-CET PYQ here reduces to the same first move — find the foot of the perpendicular by writing a parametric point and forcing perpendicularity. Once you have the foot, the question is just choosing what to do with it: report it (foot questions), reflect through it as 2F − P (mirror-image questions), or skip it entirely and dot-product (projection questions). Across the 16 PYQs the mix runs MODERATE-to-HARD, and several appear two or three times across different papers (the (5,−1,4)/(4,−1,3)-on-x+y+z=7 projection alone shows up three times) — so the patterns are stable and high-yield. Master the foot-finding routine first; image and projection are one extra line each on top of it.

Concept 1 of 7

Foot of the perpendicular from a point to a line

Intuition

Walk along the line with a parameter λ\lambda: every point on it is a⃗+λd⃗\vec{a} + \lambda\vec{d}. The foot FF is the one special point where the segment from your external point PP meets the line at a right angle. So you have ONE unknown (λ\lambda) and ONE condition (PF→⋅d⃗=0\overrightarrow{PF}\cdot\vec{d} = 0) — solve, substitute, done.

Definition

Write the line as r⃗=a⃗+λd⃗\vec{r} = \vec{a} + \lambda\vec{d} (or in symmetric form x−x0a=y−y0b=z−z0c=λ\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} = \lambda). Then:

  • Parametric foot: the foot is F=a⃗+λd⃗F = \vec{a} + \lambda\vec{d}, a single point depending on λ\lambda.
  • Perpendicularity condition: PF→⋅d⃗=0\overrightarrow{PF}\cdot\vec{d} = 0, where PF→=F−P\overrightarrow{PF} = F - P. This is **one linear equation in λ\lambda** — solve it.
  • Substitute back the λ\lambda you found into F=a⃗+λd⃗F = \vec{a} + \lambda\vec{d} to get the coordinates.

Foot on a line

F=a⃗+λd⃗,PF→⋅d⃗=0  ⇒  λF = \vec{a} + \lambda\vec{d}, \qquad \overrightarrow{PF}\cdot\vec{d} = 0 \;\Rightarrow\; \lambda
  • a⃗\vec{a}a fixed point on the line
  • d⃗\vec{d}direction vector of the line
  • PPthe external point
  • PF→\overrightarrow{PF}F−PF - P, must be perpendicular to d⃗\vec{d}
P(x₀,y₀)ax+by+c=0dd = |ax₀+by₀+c| / √(a²+b²)

Worked example

Find the foot of the perpendicular from P(1,0,2)P(1,0,2) to the line r⃗=(2i^+3k^)+λ(i^+2j^+2k^)\vec{r} = (2\hat{i} + 3\hat{k}) + \lambda(\hat{i} + 2\hat{j} + 2\hat{k}).
  1. General point on the line: F=(2+λ,  2λ,  3+2λ)F = (2+\lambda,\; 2\lambda,\; 3+2\lambda).
  2. PF→=F−P=(1+λ,  2λ,  1+2λ)\overrightarrow{PF} = F - P = (1+\lambda,\; 2\lambda,\; 1+2\lambda).
  3. Set PF→⋅d⃗=0\overrightarrow{PF}\cdot\vec{d} = 0 with d⃗=(1,2,2)\vec{d} = (1,2,2): (1+λ)+2(2λ)+2(1+2λ)=0⇒3+9λ=0⇒λ=−13(1+\lambda) + 2(2\lambda) + 2(1+2\lambda) = 0\Rightarrow 3 + 9\lambda = 0\Rightarrow \lambda = -\tfrac{1}{3}.
  4. Substitute: F=(2−13,  −23,  3−23)=(53,  −23,  73)F = \left(2-\tfrac{1}{3},\; -\tfrac{2}{3},\; 3-\tfrac{2}{3}\right) = \left(\tfrac{5}{3},\; -\tfrac{2}{3},\; \tfrac{7}{3}\right).
Answer:F=(53,  −23,  73)F = \left(\dfrac{5}{3},\; -\dfrac{2}{3},\; \dfrac{7}{3}\right)
Practice this conceptself-check · 4 quick reps

Try it yourself

Find the foot of the perpendicular from (1,2,3)(1,2,3) to the line r⃗=(6i^+7j^+7k^)+λ(3i^+2j^−2k^)\vec{r} = (6\hat{i} + 7\hat{j} + 7\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 2\hat{k}).

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    How many unknowns and conditions does foot-on-a-line have?
  2. 2.
    Line point (λ, 2λ, 0)(\lambda,\,2\lambda,\,0), d⃗=(1,2,0)\vec{d}=(1,2,0), P=(0,0,0)P=(0,0,0). What is λ\lambda?
  3. 3.
    What direction is PF→\overrightarrow{PF} perpendicular to at the foot?
  4. 4.
    After solving for λ\lambda, what is the last step?

From the bank · past-year question

Example 1Line and PlaneMODERATE
The foot of the perpendicular from the point (1,2,3)(1,2,3) on the line r⃗=(6i^+7j^+7k^)+λ(3i^+2j^−2k^)\vec{r}=(6\hat{i}+7\hat{j}+7\hat{k})+\lambda(3\hat{i}+2\hat{j}-2\hat{k}) has the co-ordinates.

[Q130 · 13th May Shift 2 · 2024]

Perpendicularity is PF→⋅d⃗=0\overrightarrow{PF}\cdot\vec{d}=0, NOT PF→=d⃗\overrightarrow{PF} = \vec{d}

You only need the dot product of the connecting segment with the line's direction to vanish — that gives one equation in one unknown. Do not try to force the whole vector PF→\overrightarrow{PF} to equal or be parallel to anything.

Use the symmetric form's parameter consistently

For x+35=y+12=z+43=λ\frac{x+3}{5} = \frac{y+1}{2} = \frac{z+4}{3} = \lambda, the point is (5λ−3, 2λ−1, 3λ−4)(5\lambda-3,\,2\lambda-1,\,3\lambda-4) — the fixed point comes from setting each numerator to 0, the multipliers (5,2,3)(5,2,3) are the direction. Mixing up which is which scrambles the sign of every coordinate (option B/C/D in these PYQs are exactly the sign-flipped traps).

Don't forget to substitute back

Finding λ\lambda is the middle of the problem, not the end. The answer is the coordinate a⃗+λd⃗\vec{a} + \lambda\vec{d}; a half-finished solution that reports λ\lambda itself is a guaranteed wrong option.
Drill 2 more on foot of the perpendicular from a point to a line

Concept 2 of 7

Foot of the perpendicular from a point to a plane

Intuition

Drop straight down from PP onto the plane — you travel along the plane's NORMAL. So shoot a line out of PP in the direction of the normal n⃗\vec{n}, march by a parameter tt, and stop the instant you hit the plane. The point where that line pierces the plane is the foot.

Definition

For a plane ax+by+cz+d=0ax + by + cz + d = 0 the normal is n⃗=(a,b,c)\vec{n} = (a,b,c). From P(x1,y1,z1)P(x_1,y_1,z_1):

  • Normal-parametric line: (x,y,z)=(x1+at,  y1+bt,  z1+ct)(x,y,z) = (x_1 + at,\; y_1 + bt,\; z_1 + ct).
  • Hit the plane: substitute these into ax+by+cz+d=0ax + by + cz + d = 0 — one linear equation in tt. Solve for tt.
  • Foot: put that tt back into the parametric point.

Compactly, t=−ax1+by1+cz1+da2+b2+c2t = -\dfrac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}, and F=P+t n⃗F = P + t\,\vec{n}.

Foot on a plane

F=P+t n⃗,t=−ax1+by1+cz1+da2+b2+c2F = P + t\,\vec{n}, \qquad t = -\dfrac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}
  • n⃗=(a,b,c)\vec{n} = (a,b,c)normal to the plane ax+by+cz+d=0ax+by+cz+d=0
  • tthow far along the normal to reach the plane
  • FFfoot of the perpendicular on the plane

Diagram · plane, normal & distance from origin (drag to rotate)

n⃗ON

Shortest path from O to the plane runs along the normal to the foot N; its length is |d| / √(a²+b²+c²).

Worked example

Find the foot of the perpendicular from P(−1,1,2)P(-1,1,2) to the plane 2x−3y+z−11=02x - 3y + z - 11 = 0.
  1. Normal n⃗=(2,−3,1)\vec{n} = (2,-3,1). Line from PP: (x,y,z)=(−1+2t,  1−3t,  2+t)(x,y,z) = (-1+2t,\; 1-3t,\; 2+t).
  2. Substitute into the plane: 2(−1+2t)−3(1−3t)+(2+t)−11=02(-1+2t) - 3(1-3t) + (2+t) - 11 = 0.
  3. Simplify: −2+4t−3+9t+2+t−11=14t−14=0⇒t=1-2+4t - 3+9t + 2+t - 11 = 14t - 14 = 0\Rightarrow t = 1.
  4. Foot: (−1+2,  1−3,  2+1)=(1,−2,3)(-1+2,\; 1-3,\; 2+1) = (1,-2,3).
Answer:F=(1,−2,3)F = (1,-2,3)
Practice this conceptself-check · 4 quick reps

Try it yourself

Find the foot of the perpendicular from P(1,1,1)P(1,1,1) to the plane x+2y+2z−9=0x + 2y + 2z - 9 = 0.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Which direction do you travel from PP to reach the foot on a plane?
  2. 2.
    Plane x+y+z=0x+y+z=0, P=(1,1,1)P=(1,1,1). Find tt.
  3. 3.
    How do you find tt once the normal line is written?
  4. 4.
    Foot formula in one line?

From the bank · past-year question

Example 2Line and PlaneMODERATE
The coordinates of the foot of the perpendicular drawn from a point P(−1,1,2)P( - 1,1,2) to the plane 2x−3y+z−11=02x- 3y+z- 11 = 0 are

[Q135 · 19 April Shift II · 2025]

Carry the constant dd with its correct sign

Write the plane as ax+by+cz+d=0ax+by+cz+d=0 first. For 2x−3y+z=112x-3y+z = 11 that is d=−11d = -11, so t=−(⋯ )−1114t = -\frac{(\cdots) - 11}{14}. Dropping the sign of dd is the single most common foot-on-plane arithmetic slip.

Divide by a2+b2+c2a^2+b^2+c^2, not by a2+b2+c2\sqrt{a^2+b^2+c^2}

The parameter tt uses ∣n⃗∣2|\vec{n}|^2 in the denominator (the distance formula's \sqrt{} appears only for actual distances, not for tt). For n⃗=(2,−3,1)\vec{n} = (2,-3,1) use 1414, never 14\sqrt{14}.
Drill 1 more on foot of the perpendicular from a point to a plane

Concept 3 of 7

Mirror image of a point in a plane

Intuition

The plane is a mirror. The foot FF is exactly halfway between PP and its image P′P' — so once you have the foot, the image is just PP reflected through FF: step the same distance past the mirror. Since FF is the midpoint of PP′PP', P′=2F−PP' = 2F - P.

Definition

Reflecting a point in a plane is foot-on-a-plane plus one reflection step:

  • Step 1 — find the foot F=P+t n⃗F = P + t\,\vec{n} using t=−ax1+by1+cz1+da2+b2+c2t = -\dfrac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2} (Concept 2).
  • Step 2 — reflect: FF is the midpoint of PP and P′P', so P′=2F−PP' = 2F - P.

A useful shortcut: since P′=2F−P=P+2t n⃗P' = 2F - P = P + 2t\,\vec{n}, the image is reached by going twice as far along the normal as the foot.

Image in a plane

P′=2F−P=P+2t n⃗,t=−ax1+by1+cz1+da2+b2+c2P' = 2F - P = P + 2t\,\vec{n}, \qquad t = -\dfrac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}
  • FFfoot of perpendicular (midpoint of PP′PP')
  • P′P'mirror image of PP in the plane
  • 2t n⃗2t\,\vec{n}twice the foot's displacement along the normal
y = xf: 2xf⁻¹: x/2(a, b) on f ⟺ (b, a) on f⁻¹

Worked example

Find the mirror image of P(2,4,−1)P(2,4,-1) in the plane x−y+2z−2=0x - y + 2z - 2 = 0, and hence a+b+ca+b+c where the image is (a,b,c)(a,b,c).
  1. Normal n⃗=(1,−1,2)\vec{n} = (1,-1,2), ∣n⃗∣2=6|\vec{n}|^2 = 6. Line: (2+t,  4−t,  −1+2t)(2+t,\; 4-t,\; -1+2t).
  2. Substitute into the plane: (2+t)−(4−t)+2(−1+2t)−2=6t−6=0⇒t=1(2+t) - (4-t) + 2(-1+2t) - 2 = 6t - 6 = 0\Rightarrow t = 1.
  3. Foot F=(3,3,1)F = (3,3,1).
  4. Image P′=2F−P=(6−2,  6−4,  2−(−1))=(4,2,3)P' = 2F - P = (6-2,\; 6-4,\; 2-(-1)) = (4,2,3). So a+b+c=9a+b+c = 9.
Answer:Image (4,2,3)(4,2,3), a+b+c=9a+b+c = 9
Practice this conceptself-check · 4 quick reps

Try it yourself

Find the image of R(2,1,6)R(2,1,6) in the plane x+y−2z−3=0x + y - 2z - 3 = 0.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    If FF is the foot, the image is P′=?P' = ?
  2. 2.
    What point is FF relative to PP and P′P'?
  3. 3.
    Foot is (3,3,1)(3,3,1), P=(2,4,−1)P=(2,4,-1). Image?
  4. 4.
    In terms of the normal step t n⃗t\,\vec{n}, the image is reached by going how far?

From the bank · past-year question

Example 3Line and PlaneHARD
The mirror image of P(2,4,−1)P(2,4,-1) in the plane x−y+2z−2=0x-y+2z-2=0 is (a,b,c)(a,b,c), then the value of a+b+ca+b+c is

[Q148 · 14th May Shift 2 · 2024]

Image is 2F−P2F - P, the foot is only halfway

A very common error is to report the FOOT as the answer to an image question. The foot is the midpoint; you must step the same distance again: P′=2F−PP' = 2F - P. The foot's coordinates are usually one of the distractor options.

2F−P2F - P, not F−2PF - 2P or 2P−F2P - F

From midpoint F=P+P′2F = \tfrac{P + P'}{2} you solve P′=2F−PP' = 2F - P — twice the foot minus the original point. Swapping the roles of FF and PP gives a reflection on the wrong side.
Drill 2 more on mirror image of a point in a plane

Concept 4 of 7

Mirror image of a point in a line

Intuition

Same midpoint idea as the plane case, but now the mirror is a LINE. First find the foot FF on the line (Concept 1), then reflect: P′=2F−PP' = 2F - P. The HARD MHT-CET twist runs it backward — you're TOLD the image and asked for an unknown coordinate aa or bb — but the same relation F=P+P′2F = \tfrac{P + P'}{2} lying on the line is all you need.

Definition

To reflect PP in a line r⃗=a⃗+λd⃗\vec{r} = \vec{a} + \lambda\vec{d}:

  • Step 1 — foot: F=a⃗+λd⃗F = \vec{a} + \lambda\vec{d} with PF→⋅d⃗=0\overrightarrow{PF}\cdot\vec{d} = 0 (Concept 1).
  • Step 2 — reflect: P′=2F−PP' = 2F - P.

**Backward ("find a,ba,b") variant:** you are given PP and its image P′P' in terms of unknowns. Two facts pin the unknowns down:

  • the midpoint P+P′2\tfrac{P + P'}{2} lies on the line (its coordinates satisfy the symmetric equation), and
  • the segment PP′→\overrightarrow{PP'} is perpendicular to d⃗\vec{d} (i.e. PP′→⋅d⃗=0\overrightarrow{PP'}\cdot\vec{d} = 0).

Solve the resulting equations for the unknowns.

Image in a line (forward and backward)

P′=2F−P;midpoint P+P′2∈line,    PP′→⋅d⃗=0P' = 2F - P; \qquad \text{midpoint } \tfrac{P+P'}{2} \in \text{line}, \;\; \overrightarrow{PP'}\cdot\vec{d} = 0
  • FFfoot of perpendicular on the line
  • P′P'mirror image of PP in the line
  • PP′→\overrightarrow{PP'}P′−PP' - P, perpendicular to the line direction

Worked example

Find the mirror image of P(0,0,0)P(0,0,0) in the line r⃗=λ(i^+j^+k^)\vec{r} = \lambda(\hat{i} + \hat{j} + \hat{k}).
  1. Foot: F=(λ,λ,λ)F = (\lambda,\lambda,\lambda), PF→=(λ,λ,λ)\overrightarrow{PF} = (\lambda,\lambda,\lambda).
  2. But PP is the origin and the line passes through the origin, so PF→⋅(1,1,1)=3λ=0⇒λ=0\overrightarrow{PF}\cdot(1,1,1) = 3\lambda = 0\Rightarrow \lambda = 0; foot F=(0,0,0)F = (0,0,0).
  3. Image P′=2F−P=(0,0,0)P' = 2F - P = (0,0,0) — the point is on the line, so it is its own image.
Answer:P′=(0,0,0)P' = (0,0,0) (the point lies on the line)
Practice this conceptself-check · 4 quick reps

Try it yourself

If the mirror image of P(a,6,9)P(a,6,9) in the line x−37=y−25=z−1−9\frac{x-3}{7} = \frac{y-2}{5} = \frac{z-1}{-9} is (20,b,−a−9)(20,b,-a-9), find ∣a+b∣|a+b|.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Image of PP in a line, given foot FF?
  2. 2.
    In the backward variant, the midpoint of PP′PP' must satisfy what?
  3. 3.
    What is perpendicular to the line direction in the backward variant?
  4. 4.
    A point that lies on the line is its own ____ in that line.

From the bank · past-year question

Example 4Line and PlaneHARD
Let a,b∈Ra,b\in\mathbb{R}. If the mirror image of the point P(a,6,9)P(a,6,9) w.r.t. line x−37=y−25=z−1−9\frac{x-3}{7}=\frac{y-2}{5}=\frac{z-1}{-9} is (20,b,−a−9)(20,b,-a-9), then ∣a+b∣|a+b| is equal to

[Q130 · 4th May Shift 1 · 2023]

Backward problems use TWO conditions, not one

When the image is given and a coordinate is unknown, one equation (midpoint on line) is not enough if there are two unknowns. Add PP′→⊥d⃗\overrightarrow{PP'}\perp\vec{d} (or a second midpoint-ratio equation) to pin both aa and bb.

Reflect in the LINE, not the line's fixed point

The mirror is the entire line, so the foot FF is the nearest point on the line to PP — found via perpendicularity — not the arbitrary point a⃗\vec{a} printed in the equation. Reflecting through a⃗\vec{a} gives the wrong image.

Concept 5 of 7

Image of a line in a plane (and planes through an image)

Intuition

Reflecting a whole line in a plane sounds bigger than it is. A line is "a point plus a direction." When the line is parallel to the plane (its direction is perpendicular to the normal), the reflection PRESERVES the direction — so you only have to reflect ONE point on the line and reattach the same direction.

Definition

To reflect a line x−x0p=y−y0q=z−z0r\frac{x-x_0}{p} = \frac{y-y_0}{q} = \frac{z-z_0}{r} in a plane with normal n⃗\vec{n}:

  • **Check direction ⊥\perp normal:** if d⃗⋅n⃗=0\vec{d}\cdot\vec{n} = 0 the line is parallel to the plane and its direction is unchanged by the reflection.
  • Reflect one point: take any point (x0,y0,z0)(x_0,y_0,z_0) on the line and find its image P′P' using Concept 3 (P′=2F−PP' = 2F - P).
  • Reassemble: the image line is x−x0′p=y−y0′q=z−z0′r\dfrac{x - x_0'}{p} = \dfrac{y - y_0'}{q} = \dfrac{z - z_0'}{r} — image point, same direction ratios (p,q,r)(p,q,r).

A close cousin asks for a plane through an image point containing a given line: reflect the point (Concept 3), then build the plane through that image point and the line.

Image line — reflect point, preserve direction

d⃗⋅n⃗=0  ⇒  line∥plane,image line={P′,  d⃗}\vec{d}\cdot\vec{n} = 0 \;\Rightarrow\; \text{line} \parallel \text{plane}, \quad \text{image line} = \{P',\; \vec{d}\}
  • d⃗=(p,q,r)\vec{d} = (p,q,r)direction of the original line, preserved if d⃗⋅n⃗=0\vec{d}\cdot\vec{n}=0
  • P′P'image of a point on the line (via 2F−P2F - P)
  • n⃗\vec{n}normal of the mirror plane

Worked example

Find the image of the line x−13=y−31=z−4−5\frac{x-1}{3} = \frac{y-3}{1} = \frac{z-4}{-5} in the plane 2x−y+z+3=02x - y + z + 3 = 0.
  1. Check direction vs normal: d⃗=(3,1,−5)\vec{d} = (3,1,-5), n⃗=(2,−1,1)\vec{n} = (2,-1,1); d⃗⋅n⃗=6−1−5=0\vec{d}\cdot\vec{n} = 6 - 1 - 5 = 0. Line is parallel to the plane → direction preserved.
  2. Reflect the point P(1,3,4)P(1,3,4): t=−2(1)−3+4+322+(−1)2+12=−66=−1t = -\dfrac{2(1) - 3 + 4 + 3}{2^2 + (-1)^2 + 1^2} = -\dfrac{6}{6} = -1. Foot F=(1,3,4)+(−1)(2,−1,1)=(−1,4,3)F = (1,3,4) + (-1)(2,-1,1) = (-1,4,3).
  3. Image point P′=2F−P=(−2−1,  8−3,  6−4)=(−3,5,2)P' = 2F - P = (-2-1,\; 8-3,\; 6-4) = (-3,5,2).
  4. Reassemble with the same direction (3,1,−5)(3,1,-5): image line x+33=y−51=z−2−5\dfrac{x+3}{3} = \dfrac{y-5}{1} = \dfrac{z-2}{-5}.
Answer:x+33=y−51=z−2−5\dfrac{x+3}{3} = \dfrac{y-5}{1} = \dfrac{z-2}{-5}
Practice this conceptself-check · 4 quick reps

Try it yourself

Let PP be the image of (3,1,7)(3,1,7) in the plane x−y+z=3x - y + z = 3. Find the equation of the plane through PP containing the line x1=y2=z1\frac{x}{1} = \frac{y}{2} = \frac{z}{1}.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    When is a line's direction unchanged by reflection in a plane?
  2. 2.
    To reflect a whole parallel line, how many points must you reflect?
  3. 3.
    Direction (3,1,−5)(3,1,-5), normal (2,−1,1)(2,-1,1): is the line parallel to the plane?
  4. 4.
    Image line = image point plus what?

From the bank · past-year question

Example 5Line and PlaneHARD
The image of the line x−13=y−31=z−4−5\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5} in the plane 2x−y+z+3=02x-y+z+3=0 is the line

[Q138 · 3rd May 2nd Shift · 2023]

Preserve the direction ratios — don't negate them

For a line parallel to the plane, the image line keeps the SAME (p,q,r)(p,q,r). Distractors flip the signs of the direction (and shift the image point) — only the point moves under reflection, the direction stays put. Verify d⃗⋅n⃗=0\vec{d}\cdot\vec{n}=0 before relying on this.

Reflect a point ON the line, then reattach the direction

Don't try to reflect the direction vector through the plane separately. Reflect a concrete point (the numerators give you one), get P′=2F−PP' = 2F - P, and the image line is P′P' with the original direction.
Drill 1 more on image of a line in a plane (and planes through an image)

Concept 6 of 7

Projection of a segment onto a line

Intuition

Shine a light perpendicular to a line and measure the shadow a segment casts on it. That shadow length is the segment's component along the line's direction — a single dot product divided by the direction's length.

Definition

The projection (length of the shadow) of the segment AB→\overrightarrow{AB} onto a line with direction d⃗\vec{d} is

proj=∣AB→⋅d⃗∣∣d⃗∣.\text{proj} = \frac{|\overrightarrow{AB}\cdot\vec{d}|}{|\vec{d}|}.

  • Compute AB→=B−A\overrightarrow{AB} = B - A.
  • Dot it with the line's direction ratios d⃗\vec{d}.
  • Divide by ∣d⃗∣=a2+b2+c2|\vec{d}| = \sqrt{a^2+b^2+c^2}; take the absolute value (a length is non-negative).

Projection onto a line

proj=∣AB→⋅d⃗∣∣d⃗∣\text{proj} = \dfrac{|\overrightarrow{AB}\cdot\vec{d}|}{|\vec{d}|}
  • AB→\overrightarrow{AB}B−AB - A, the segment vector
  • d⃗\vec{d}direction ratios of the line
  • ∣d⃗∣|\vec{d}|a2+b2+c2\sqrt{a^2+b^2+c^2}

Worked example

Find the projection of the segment joining A(1,2,3)A(1,2,3) and B(4,4,9)B(4,4,9) on the line with direction ratios (2,1,2)(2,1,2).
  1. AB→=(3,2,6)\overrightarrow{AB} = (3,2,6).
  2. Dot with d⃗=(2,1,2)\vec{d} = (2,1,2): 3(2)+2(1)+6(2)=6+2+12=203(2) + 2(1) + 6(2) = 6 + 2 + 12 = 20.
  3. ∣d⃗∣=4+1+4=3|\vec{d}| = \sqrt{4+1+4} = 3.
  4. Projection =∣20∣3=203= \dfrac{|20|}{3} = \dfrac{20}{3}.
Answer:203\dfrac{20}{3} units
Practice this conceptself-check · 4 quick reps

Try it yourself

Find the projection of the segment joining (2,1,−3)(2,1,-3) and (−1,0,2)(-1,0,2) on the line with direction ratios (3,2,6)(3,2,6).

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Projection of AB→\overrightarrow{AB} on direction d⃗\vec{d}?
  2. 2.
    AB→=(1,2,2)\overrightarrow{AB} = (1,2,2), d⃗=(2,2,1)\vec{d} = (2,2,1). Projection?
  3. 3.
    Why take absolute value?
  4. 4.
    Divide the dot product by ∣d⃗∣|\vec{d}| or ∣d⃗∣2|\vec{d}|^2?

From the bank · past-year question

Example 6Line and PlaneMODERATE
The projection of the line segment joining the points (2,1,−3)(2, 1, -3) and (−1,0,2)(-1, 0, 2) on the line whose direction ratios are 3,2,63, 2, 6 is

[Shift || · 2025]

Divide by ∣d⃗∣|\vec{d}|, not ∣d⃗∣2|\vec{d}|^2

The projection LENGTH uses ∣d⃗∣|\vec{d}| (one power) in the denominator. Dividing by ∣d⃗∣2|\vec{d}|^2 gives the scalar multiplier for the projection VECTOR, not its length — a frequent mix-up with vector projection.

AB→=B−A\overrightarrow{AB} = B - A — direction matters inside the dot product

Get the head-minus-tail right. The magnitude is unaffected by swapping AA and BB (the absolute value cleans up the sign), but a component mistake in B−AB - A changes the dot product itself.
Drill 1 more on projection of a segment onto a line

Concept 7 of 7

Projection of a segment onto a plane

Intuition

Now the shadow falls on a PLANE (light shining straight down the normal). The segment splits into a part along the normal (lost in the shadow) and a part in the plane (the shadow itself). By Pythagoras, the in-plane shadow is the full length with the along-normal piece removed: ∣AB→∣2−(normal component)2\sqrt{|\overrightarrow{AB}|^2 - (\text{normal component})^2}.

Definition

The **projection of AB→\overrightarrow{AB} onto a plane** with unit normal n^\hat{n} is the part of AB→\overrightarrow{AB} lying in the plane. Its length is

projplane=∣AB→∣2−(AB→⋅n^)2.\text{proj}_{\text{plane}} = \sqrt{|\overrightarrow{AB}|^2 - (\overrightarrow{AB}\cdot\hat{n})^2}.

  • ∣AB→∣2|\overrightarrow{AB}|^2 is the full squared length.
  • AB→⋅n^=AB→⋅n⃗∣n⃗∣\overrightarrow{AB}\cdot\hat{n} = \dfrac{\overrightarrow{AB}\cdot\vec{n}}{|\vec{n}|} is the component ALONG the normal (the bit that gets flattened away).
  • Subtract its square and take the root — the in-plane shadow.

Projection onto a plane

projplane=∣AB→∣2−(AB→⋅n⃗∣n⃗∣)2\text{proj}_{\text{plane}} = \sqrt{|\overrightarrow{AB}|^2 - \left(\dfrac{\overrightarrow{AB}\cdot\vec{n}}{|\vec{n}|}\right)^2}
  • AB→\overrightarrow{AB}the segment vector B−AB - A
  • n⃗\vec{n}normal to the plane
  • AB→⋅n⃗∣n⃗∣\dfrac{\overrightarrow{AB}\cdot\vec{n}}{|\vec{n}|}the along-normal component (removed)

Diagram · plane, normal & distance from origin (drag to rotate)

n⃗ON

Shortest path from O to the plane runs along the normal to the foot N; its length is |d| / √(a²+b²+c²).

Worked example

Find the length of the projection of the segment joining (5,−1,4)(5,-1,4) and (4,−1,3)(4,-1,3) on the plane x+y+z=7x + y + z = 7.
  1. AB→=(4−5,  −1+1,  3−4)=(−1,0,−1)\overrightarrow{AB} = (4-5,\; -1+1,\; 3-4) = (-1,0,-1), so ∣AB→∣2=1+0+1=2|\overrightarrow{AB}|^2 = 1 + 0 + 1 = 2.
  2. Normal n⃗=(1,1,1)\vec{n} = (1,1,1), ∣n⃗∣=3|\vec{n}| = \sqrt{3}. Along-normal component: AB→⋅n⃗∣n⃗∣=−1+0−13=−23\dfrac{\overrightarrow{AB}\cdot\vec{n}}{|\vec{n}|} = \dfrac{-1+0-1}{\sqrt{3}} = \dfrac{-2}{\sqrt{3}}, squared =43= \dfrac{4}{3}.
  3. Projection =2−43=23= \sqrt{2 - \dfrac{4}{3}} = \sqrt{\dfrac{2}{3}}.
Answer:23\sqrt{\dfrac{2}{3}} units
Practice this conceptself-check · 4 quick reps

Try it yourself

Find the projection of the segment joining (1,0,0)(1,0,0) and (1,1,1)(1,1,1) on the plane z=0z = 0.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Projection of AB→\overrightarrow{AB} onto a plane (formula)?
  2. 2.
    Which component of AB→\overrightarrow{AB} is removed when projecting onto a plane?
  3. 3.
    ∣AB→∣2=2|\overrightarrow{AB}|^2 = 2, normal component squared =43= \tfrac{4}{3}. Projection?
  4. 4.
    For projection onto a LINE you keep AB→⋅d⃗\overrightarrow{AB}\cdot\vec d; onto a PLANE you remove what?

From the bank · past-year question

Example 7Line and PlaneMODERATE
The length (in units) of the projection of the line segment, joining the points (5,−1,4)(5,-1,4) and (4,−1,3)(4,-1,3), on the plane x+y+z=7x+y+z=7 is

[Q136 · 12th May Shift 2 · 2024]

Plane projection SUBTRACTS the normal part; line projection KEEPS the direction part

Onto a line you want the component ALONG d⃗\vec{d}. Onto a plane you want what's LEFT after removing the component along n⃗\vec{n}. Using the line formula for a plane question (or vice-versa) is the headline trap — they are complementary pieces of the same vector.

Use the UNIT normal inside the square

The removed piece is (AB→⋅n^)2=(AB→⋅n⃗∣n⃗∣)2(\overrightarrow{AB}\cdot\hat{n})^2 = \left(\frac{\overrightarrow{AB}\cdot\vec{n}}{|\vec{n}|}\right)^2. Forgetting to divide by ∣n⃗∣|\vec{n}| over-counts the normal component. For n⃗=(1,1,1)\vec{n}=(1,1,1), divide the dot product by 3\sqrt{3} before squaring.

Answer is 2/3\sqrt{2/3}, not 2/32/3

Take the final square root. 23\sqrt{\frac{2}{3}} and 23\frac{2}{3} are both offered as options in this PYQ — the un-rooted value is the classic distractor.
Drill 2 more on projection of a segment onto a plane

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (7)

  • Foot of the perpendicular from a point to a line

    Foot on a line

    F=a⃗+λd⃗,PF→⋅d⃗=0  ⇒  λF = \vec{a} + \lambda\vec{d}, \qquad \overrightarrow{PF}\cdot\vec{d} = 0 \;\Rightarrow\; \lambda
  • Foot of the perpendicular from a point to a plane

    Foot on a plane

    F=P+t n⃗,t=−ax1+by1+cz1+da2+b2+c2F = P + t\,\vec{n}, \qquad t = -\dfrac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}
  • Mirror image of a point in a plane

    Image in a plane

    P′=2F−P=P+2t n⃗,t=−ax1+by1+cz1+da2+b2+c2P' = 2F - P = P + 2t\,\vec{n}, \qquad t = -\dfrac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}
  • Mirror image of a point in a line

    Image in a line (forward and backward)

    P′=2F−P;midpoint P+P′2∈line,    PP′→⋅d⃗=0P' = 2F - P; \qquad \text{midpoint } \tfrac{P+P'}{2} \in \text{line}, \;\; \overrightarrow{PP'}\cdot\vec{d} = 0
  • Image of a line in a plane (and planes through an image)

    Image line — reflect point, preserve direction

    d⃗⋅n⃗=0  ⇒  line∥plane,image line={P′,  d⃗}\vec{d}\cdot\vec{n} = 0 \;\Rightarrow\; \text{line} \parallel \text{plane}, \quad \text{image line} = \{P',\; \vec{d}\}
  • Projection of a segment onto a line

    Projection onto a line

    proj=∣AB→⋅d⃗∣∣d⃗∣\text{proj} = \dfrac{|\overrightarrow{AB}\cdot\vec{d}|}{|\vec{d}|}
  • Projection of a segment onto a plane

    Projection onto a plane

    projplane=∣AB→∣2−(AB→⋅n⃗∣n⃗∣)2\text{proj}_{\text{plane}} = \sqrt{|\overrightarrow{AB}|^2 - \left(\dfrac{\overrightarrow{AB}\cdot\vec{n}}{|\vec{n}|}\right)^2}

Watch out for (16)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Line and PlaneMODERATE
The co-ordinates of the foot of the perpendicular from the point (0,2,3)(0,2,3) on the line x+35=y+12=z+43\frac{x+3}{5} = \frac{y+1}{2} = \frac{z+4}{3} is

[Q105 · 16th May Shift 2 · 2023]

Example 2Line and PlaneMODERATE
The coordinates of the foot of the perpendicular drawn from a point P(-1, 1, 2) to the plane 2x−3y+z−11=02x - 3y + z - 11 = 0

[Shift || · 2025]

Example 3Line and PlaneHARD
Let P be a plane passing through the points (2,1,0),(4,1,1)(2,1,0),(4,1,1) and (5,0,1)(5,0,1) and R be the point (2,1,6)(2,1,6). Then image of R in the plane P is

[Q108 · 4th May Shift 2 · 2023]

Example 4Line and PlaneHARD
Let P be the image of the point (3,1,7)(3,1,7) with respect to the plane x−y+z=3x-y+z=3. Then the equation of the plane passing through P and containing the straight line x1=y2=z1\frac{x}{1}=\frac{y}{2}=\frac{z}{1} is

[Q144 · 2nd May Shift 2 · 2023]

Example 5Line and PlaneMODERATE
The projection of the line segment joining the points (2,1,−3)(2,1, - 3) and (−1,0,2)( - 1,0,2) on the line whose direction ratios are 3,2,63,2,6 is

[Q114 · 19 April Shift II · 2025]

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