MathsCBSE Class 12 · Three Dimensional Geometry

Foot of the Perpendicular from a Point to a Line, and the Image of a Point (Class 12)

Updated 15 October 2026 · Fact-checked

Write the general point P(λ) of the line. The vector from the given point A to P(λ) must be perpendicular to the line's direction b, so set AP · b = 0 and solve for λ; that P is the foot N. The perpendicular distance is |AN|, and the image of A in the line is 2N − A.

Only at the foot N is AP perpendicular to direction b; the image A′ lies the same distance beyond N, so A′ = 2N − A.

Illustration drawn for QuestionBuddy.

Understand Foot of Perpendicular and Image of a Point

Questions in this topic ask for three related things: the foot of the perpendicular $N$ from a point $A$ to a line $L$ , the perpendicular distance $A N$ (the shortest distance from $A$ to $L$ ), and the image (reflection) $A^{'}$ of $A$ in $L$ .

The general-point method (always works).

1. Write the line as $\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c} = λ$ , so any point on it is $P (x_{1} + aλ, y_{1} + bλ, z_{1} + c λ)$ .
2. Form $A P = P - A$ in terms of $λ$ .
3. $A P$ is perpendicular to the line exactly when $A P \cdot (a, b, c) = 0$ . This is a linear equation in $λ$ ; solve it.
4. Substitute $λ$ to get the foot $N$ .

Distance and image. The perpendicular distance is $∣ A N ∣$ . Since $N$ is the midpoint of $A$ and its image $A^{'}$ , we get $A^{'} = 2 N - A$ , i.e. $x^{'} = 2 x_{N} - x_{A}$ , and similarly for $y$ and $z$ .

A formula check. For the line $r = a + λ b$ , the distance from $A$ is
$d = \frac{∣ ( A - a ) \times b ∣}{∣ b ∣}$ ,
and the foot is $N = a + \frac{( A - a ) \cdot b}{∣ b ∣ ^{2}} b$ . These are handy for checking, but board answers usually expect the general-point method.

Special lines. For a coordinate axis the foot is immediate: from $(p, q, r)$ to the x-axis it is $(p, 0, 0)$ and the distance is $q^{2} + r^{2}$ ; to the z-axis it is $(0, 0, r)$ with distance $p^{2} + q^{2}$ .

Uses. The same method gives the altitude of a triangle from a vertex (the foot on the opposite side), the equation of the perpendicular from $A$ to $L$ (the line through $A$ and $N$ ), and checks like 'is this point the reflection of that one'. Checking your foot is easy: $N$ must satisfy the line's equation and $A N \cdot b$ must be 0.

This is among the highest-scoring 3D topics: image and foot-of-perpendicular questions appeared as 5-mark questions in the 2024, 2025 and 2026 board papers (NCERT Section 11.3, through the general point).

Foot of Perpendicular and Image of a Point: formulas to remember

General point: $P (λ) = (x_{1} + aλ, y_{1} + bλ, z_{1} + c λ)$; From $\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c} = λ$ .
Perpendicularity condition: $A P \cdot (a \hat{i} + b \hat{j} + c \hat{k}) = 0$; Solve for λ to locate the foot.
Image of A: $A^{'} = 2 N - A$; N is the midpoint of $A A^{'}$ .
Perpendicular distance: $d = ∣ A N ∣ = \frac{∣ ( A - a ) \times b ∣}{∣ b ∣}$; The cross-product form is a quick check.
Foot by projection: $N = a + \frac{( A - a ) \cdot b}{∣ b ∣ ^{2}} b$; Same answer as the general-point method.
Distance from the axes: $(p, q, r) : d_{x} = q^{2} + r^{2}, d_{y} = p^{2} + r^{2}, d_{z} = p^{2} + q^{2}$; Feet $(p, 0, 0)$ , $(0, q, 0)$ , $(0, 0, r)$ .

How to solve Foot of Perpendicular and Image of a Point questions

The standard 5-mark layout for foot, distance and image questions.

1Rewrite the line in standard Cartesian form and put each fraction equal to $λ$ ; write the general point P.
2Write $A P$ as $(P_{x} - A_{x}, P_{y} - A_{y}, P_{z} - A_{z})$ in terms of $λ$ .
3Set $A P \cdot b = 0$ , using the direction ratios of the line, and solve for $λ$ .
4Substitute $λ$ into P to get the foot N; check it with $A N \cdot b = 0$ .
5Distance: $∣ A N ∣$ . Image: $A^{'} = 2 N - A$ . Equation of the perpendicular: the line through A and N.

Foot of Perpendicular and Image of a Point in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Shortest distance from a point to a line

Suppose a drone hovers at the point $(6, 8, 10)$ (in metres) near a straight power cable that runs along the x-axis.

Find the point of the cable nearest to the drone and the drone's distance from the cable.

Try it first, then show the solutionSolution

The cable is the x-axis: general point $(λ, 0, 0)$ , direction $\hat{i}$ .
$A P = (λ - 6, - 8, - 10)$ ; $A P \cdot \hat{i} = λ - 6 = 0 \Rightarrow λ = 6$ .
Foot $N (6, 0, 0)$ ; distance $= 0 + 64 + 100 = 164 \approx 12.8$ m.

Answer: Nearest point $(6, 0, 0)$ ; distance $164 \approx 12.8$ m.

Memory hook: The nearest point on a line is the foot of the perpendicular.

Foot of the perpendicular on a slanting line

Imagine a security camera at $A (1, 1, 3)$ and a straight walkway along $r = λ (\hat{i} + \hat{j})$ on the ground (in metres).

Find the point of the walkway closest to the camera and the distance.

Try it first, then show the solutionSolution

General point $P (λ, λ, 0)$ ; $A P = (λ - 1, λ - 1, - 3)$ .
$A P \cdot (1, 1, 0) = 2 λ - 2 = 0 \Rightarrow λ = 1$ ; foot $(1, 1, 0)$ .
Distance $= ∣ (0, 0, - 3) ∣ = 3$ m: the camera is directly above that point.

Answer: Closest point $(1, 1, 0)$ , at distance 3 m.

Memory hook: Set the dot product to zero, solve for λ, plug back in.

Tricks to remember Foot of Perpendicular and Image of a Point

Short memory aids. Say each one aloud once, then use it on a question the same day.

General point, dot zero, plug back
Three moves solve every foot question: write P(λ), set $A P \cdot b = 0$ , substitute λ.
Use it when: Foot, distance and image.
Image is twice the foot minus the point
$A^{'} = 2 N - A$ : the foot is halfway between a point and its reflection.
Use it when: Image questions.
Dot with the line's direction, not A's position
The dot product is with the direction ratios of the line, never with the coordinates of A.
Use it when: Avoiding the commonest error.
Check N twice
N must lie on the line, and $A N$ must be perpendicular to it.
Use it when: A 30-second check.

Foot of Perpendicular and Image of a Point: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Foot on a line through the origin · 3 marks · Short answerBoard level

Find the foot of the perpendicular from $A (2, 4, - 1)$ to the line $x = y = z$ , and the perpendicular distance.

Try it first, then show the solutionSolution

General point $P (λ, λ, λ)$ , direction $(1, 1, 1)$ ; $A P = (λ - 2, λ - 4, λ + 1)$ . (1 mark)
$A P \cdot (1, 1, 1) = 3 λ - 5 = 0 \Rightarrow λ = \frac{5}{3}$ ; foot $N (\frac{5}{3}, \frac{5}{3}, \frac{5}{3})$ . (1 mark)
$A N = (- \frac{1}{3}, - \frac{7}{3}, \frac{8}{3})$ , distance $= \frac{1 + 49 + 64}{3} = \frac{114}{3}$ . (1 mark)

Answer: $N (\frac{5}{3}, \frac{5}{3}, \frac{5}{3})$ ; distance $\frac{114}{3}$

Example 2: Foot of the perpendicular from the origin · 3 marks · Short answerBoard level

Find the foot of the perpendicular from the origin to the line through $(- 1, 2, 3)$ with direction ratios $1, - 1, 2$ , and its distance from the origin.

Try it first, then show the solutionSolution

General point $P (- 1 + λ, 2 - λ, 3 + 2 λ)$ ; $O P = P$ . (1 mark)
$P \cdot (1, - 1, 2) = - 1 + λ - 2 + λ + 6 + 4 λ = 3 + 6 λ = 0 \Rightarrow λ = - \frac{1}{2}$ . (1 mark)
Foot $(- \frac{3}{2}, \frac{5}{2}, 2)$ ; distance $= \frac{9}{4} + \frac{25}{4} + 4 = \frac{50}{4} = \frac{5}{2}$ . (1 mark)

Answer: Foot $(- \frac{3}{2}, \frac{5}{2}, 2)$ ; distance $\frac{5}{2}$

Example 3: Foot, distance and image in one question · 5 marks · Long answerHarder board level

Find the foot of the perpendicular from $A (1, 2, 3)$ to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ . Hence find the perpendicular distance and the image of A in the line.

Try it first, then show the solutionSolution

General point $P (6 + 3 λ, 7 + 2 λ, 7 - 2 λ)$ . (1 mark)
$A P = (5 + 3 λ, 5 + 2 λ, 4 - 2 λ)$ . (½ mark)
$A P \cdot (3, 2, - 2) = 15 + 9 λ + 10 + 4 λ - 8 + 4 λ = 17 + 17 λ = 0 \Rightarrow λ = - 1$ . (1½ marks)
Foot $N (3, 5, 9)$ ; $A N = (2, 3, 6)$ , distance $= 4 + 9 + 36 = 7$ . (1 mark)
Image $A^{'} = 2 N - A = (6 - 1, 10 - 2, 18 - 3) = (5, 8, 15)$ . (1 mark)

Answer: Foot $(3, 5, 9)$ ; distance 7; image $(5, 8, 15)$

Example 4: Distance by the cross-product formula · 3 marks · Short answerHarder board level

Find the perpendicular distance of $P (3, - 1, 2)$ from the line $r = (\hat{i} + \hat{j}) + λ (2 \hat{i} - \hat{j} + 2 \hat{k})$ .

Try it first, then show the solutionSolution

$A P = (3 - 1, - 1 - 1, 2 - 0) = (2, - 2, 2)$ with $A (1, 1, 0)$ on the line; $b = (2, - 1, 2)$ , $∣ b ∣ = 3$ . (1 mark)
$A P \times b = \hat{i} (- 4 + 2) - \hat{j} (4 - 4) + \hat{k} (- 2 + 4) = - 2 \hat{i} + 2 \hat{k}$ , magnitude $22$ . (1 mark)
Distance $= \frac{2 2}{3}$ . (1 mark)

Answer: $\frac{2 2}{3}$

Example 5: Altitude of a triangle in space · 5 marks · Long answerHOTS / competency

For the triangle with vertices $A (1, 2, 3)$ , $B (2, - 1, 0)$ and $C (4, 1, 2)$ , find the foot of the altitude from A on BC, the length of the altitude, and hence the area of the triangle.

Try it first, then show the solutionSolution

$B C = (2, 2, 2)$ , so take direction $(1, 1, 1)$ ; general point of BC: $P (2 + λ, - 1 + λ, λ)$ . (1 mark)
$A P = (1 + λ, - 3 + λ, λ - 3)$ ; $A P \cdot (1, 1, 1) = 3 λ - 5 = 0 \Rightarrow λ = \frac{5}{3}$ . (1 mark)
Foot $N (\frac{11}{3}, \frac{2}{3}, \frac{5}{3})$ ; $A N = (\frac{8}{3}, - \frac{4}{3}, - \frac{4}{3})$ , length $\frac{96}{3} = \frac{4 6}{3}$ . (1½ marks)
$∣ B C ∣ = 23$ , so area $= \frac{1}{2} \times 23 \times \frac{4 6}{3} = \frac{4 18}{3} = 42$ . (1 mark)
Check: $\frac{1}{2} ∣ A B \times A C ∣ = \frac{1}{2} ∣ (0, - 8, 8) ∣ = 42$ . (½ mark)

Answer: Foot $(\frac{11}{3}, \frac{2}{3}, \frac{5}{3})$ ; altitude $\frac{4 6}{3}$ ; area $42$

Example 6: Case study: a lamp post and a child · 4 marks · Case-basedHOTS / competency

Suppose a vertical lamp post stands along the line through $(2, 3, 0)$ with direction $\hat{k}$ (in metres). A child's ball is at $A (5, 7, 4)$ .
(i) Write the equation of the lamp post line. (1 mark)
(ii) Find the foot of the perpendicular from A to the post. (1 mark)
(iii) Find the distance of the ball from the post. (1 mark)
(iv) Find the image of A in the post's line. (1 mark)

Try it first, then show the solutionSolution

(i) $r = 2 \hat{i} + 3 \hat{j} + λ \hat{k}$ ; general point $(2, 3, λ)$ . (1 mark)
(ii) $A P = (- 3, - 4, λ - 4)$ ; $A P \cdot \hat{k} = λ - 4 = 0 \Rightarrow N (2, 3, 4)$ . (1 mark)
(iii) $∣ A N ∣ = 9 + 16 = 5$ m. (1 mark)
(iv) $A^{'} = 2 N - A = (4 - 5, 6 - 7, 8 - 4) = (- 1, - 1, 4)$ . (1 mark)

Answer: (i) $r = 2 \hat{i} + 3 \hat{j} + λ \hat{k}$ (ii) $(2, 3, 4)$ (iii) 5 m (iv) $(- 1, - 1, 4)$

Where students lose marks in Foot of Perpendicular and Image of a Point

The slips examiners see most often, and what to write instead.

Mistake: Dotting with the wrong vector
The perpendicularity condition uses the direction of the line.
Wrong working
$A P \cdot O A = 0$ .
Right working
$A P \cdot b = 0$ , with $b$ the line's direction ratios.
Do this: Underline the denominators of the line and use them.
Usually costs: Most of the marks.
Mistake: Image computed as A − N
N is the midpoint of $A A^{'}$ , so $A^{'} = 2 N - A$ .
Wrong working
Image $= N - A = (2, 3, 6)$ .
Right working
Image $= 2 N - A = (5, 8, 15)$ .
Do this: Use the midpoint idea: $N = \frac{A + A ^{'}}{2}$ .
Usually costs: 1 to 2 marks.
Mistake: Not rewriting the line in standard form
The general point is wrong if the numerators are not $x - x_{1}$ .
Wrong working
$\frac{1 - x}{2} = \dots$ used with point $(1, \dots)$ and ratio 2.
Right working
$\frac{x - 1}{- 2} = \dots$ : ratio $- 2$ .
Do this: Clean the line first.
Usually costs: The foot and everything after it.
Mistake: Arithmetic slips solving for λ
Expanding $A P \cdot b$ has many signs.
Wrong working
$15 + 9 λ + 10 + 4 λ + 8 - 4 λ = 0$ .
Right working
$15 + 9 λ + 10 + 4 λ - 8 + 4 λ = 0 \Rightarrow λ = - 1$ .
Do this: Write each product term with its sign, then collect.
Usually costs: The answer.

Exam tips

Foot-of-perpendicular and image questions were 5-mark questions in the 2024, 2025 and 2026 board papers; follow the four-step layout exactly.
Verify your foot by substituting into the line and checking $A N \cdot b = 0$ ; it catches most slips.
The distance can be cross-checked with $\frac{∣ ( A - a ) \times b ∣}{∣ b ∣}$ .
State the image as coordinates and, if asked, as a position vector.
Plane-based reflections are not in the 2026-27 syllabus; all questions use lines.

Practice questions on Foot of Perpendicular and Image of a Point

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The distance of the point $(3, 4, 5)$ from the z-axis is:
(a) 3
(b) 4
(c) 5
(d) $52$
Hint
Foot on the z-axis is $(0, 0, 5)$ .
Answer
(c) $9 + 16 = 5$ .
Q2 · 1 mark · Very short answerBoard level
Write the foot of the perpendicular from $(1, 2, 3)$ to the x-axis and the perpendicular distance.
Hint
On the x-axis, y = z = 0.
Answer
Foot $(1, 0, 0)$ ; distance $4 + 9 = 13$ .
Q3 · 3 marks · Short answerBoard level
Find the foot of the perpendicular from $(2, - 1, 5)$ to the line through the origin with direction ratios $1, 2, 2$ .
Hint
General point $(λ, 2 λ, 2 λ)$ .
Answer
$A P = (λ - 2, 2 λ + 1, 2 λ - 5)$ ; dot with $(1, 2, 2)$ : $9 λ - 10 = 0 \Rightarrow λ = \frac{10}{9}$ . Foot $(\frac{10}{9}, \frac{20}{9}, \frac{20}{9})$ .
Q4 · 5 marks · Long answerHarder board level
Find the image of the point $A (4, 0, 4)$ in the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{2}$ , and the distance of A from the line.
Hint
General point $(1 + 2 λ, - 1 - λ, 2 + 2 λ)$ .
Answer
$A P = (2 λ - 3, - 1 - λ, 2 λ - 2)$ ; dot with $(2, - 1, 2)$ : $9 λ - 9 = 0 \Rightarrow λ = 1$ . Foot $N (3, - 2, 4)$ . Distance $∣ A N ∣ = ∣ (- 1, - 2, 0) ∣ = 5$ . Image $2 N - A = (2, - 4, 4)$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The foot of the perpendicular from $(a, b, c)$ to the y-axis is $(0, b, 0)$ .
Reason (R): The perpendicular from a point to the y-axis meets it at the point with the same y-coordinate.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Use the general point $(0, λ, 0)$ .
Answer
(a). $A P \cdot \hat{j} = λ - b = 0$ gives $λ = b$ , which is exactly R.
Q6 · 3 marks · Short answerHarder board level
Find the distance of $(1, 1, 1)$ from the line $r = λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ .
Hint
Use $\frac{∣ A \times b ∣}{∣ b ∣}$ with the line through the origin.
Answer
$(1, 1, 1) \times (2, 3, 6) = (3, - 4, 1)$ , magnitude $26$ ; $∣ b ∣ = 7$ ; distance $\frac{26}{7}$ .
Q7 · 2 marks · Very short answerHOTS / competency
For the line and point of the previous question with image $(2, - 4, 4)$ (that is, $A (4, 0, 4)$ and the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{2}$ ), find the equation of the perpendicular from A to the line.
Hint
It passes through A and the foot $N (3, - 2, 4)$ .
Answer
Direction $A N = (- 1, - 2, 0)$ : $\frac{x - 4}{- 1} = \frac{y}{- 2}$ , $z = 4$ (equivalently $r = 4 \hat{i} + 4 \hat{k} + μ (\hat{i} + 2 \hat{j})$ ). Check: $(- 1, - 2, 0) \cdot (2, - 1, 2) = 0$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Condition for the foot N of the perpendicular from A to a line with direction $b$ ?Show answer
$A N \cdot b = 0$ , with N on the line.
2.Image of A in a line with foot N?Show answer
$2 N - A$ .
3.Distance of $(p, q, r)$ from the x-axis?Show answer
$q^{2} + r^{2}$ .
4.Cross-product formula for the distance from a point to a line?Show answer
$\frac{∣ ( A - a ) \times b ∣}{∣ b ∣}$ .
5.Foot of the perpendicular from $(2, 5, 7)$ to the z-axis?Show answer
$(0, 0, 7)$ .
6.What is the first step in every foot question?Show answer
Write the general point of the line in terms of λ.

Foot of Perpendicular and Image of a Point: frequently asked questions

How do I find the foot of the perpendicular from a point to a line?

Write the general point of the line using λ, form the vector from the given point to it, set its dot product with the line's direction ratios equal to zero, solve for λ and substitute back.

How do I find the image of a point in a line?

Find the foot N first. The image A′ satisfies N = (A + A′)/2, so A′ = 2N − A.

Is there a formula for the perpendicular distance?

Yes: |(A − a) × b|/|b| for the line r = a + λb. It is a quick check, but showing the foot and then the distance is the expected method.

Which NCERT section supports this topic?

Section 11.3 (Equation of a Line in Space) of Chapter 11 in NCERT Mathematics Part II for Class 12, using the general point of a line; the chapter's miscellaneous exercise has similar problems.

Can I use the plane method for the image?

No. Planes were removed from the 2026-27 syllabus. Use the general-point method with lines only.