---
title: "Perpendicular from point to line"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:perpendicular-from-point-line"
source: https://www.bctengineer.com/q/semester_2/mathematics/perpendicular-from-point-line
site: BCT Engineering
---

# Perpendicular from point to line

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2072 Magh, 5 marks; 2071 Bhadra, 5 marks; 2071 Magh, 5 marks; 2070 Bhadra, 5 marks; 2067 Mangsir, 5 marks (asked 5 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/perpendicular-from-point-line
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Find the length of perpendicular from the point (3,−1,11) to the line x/2 = (y−2)/3 = (z−3)/4. Also obtain the equation of the perpendicular.

## Short answer

Let foot Q(2t,2+3t,3+4t). Dot (2,3,4)·PQ=0 gives t=1, Q(2,5,7). Length = √[(2-3)²+(5+1)²+(7-11)²]=√53. Perpendicular: (x-3)/-1 = (y+1)/6 = (z-11)/-4.

## Answer

Let P = (3, −1, 11) and the line L: x/2 = (y − 2)/3 = (z − 3)/4, which passes through A(0, 2, 3) and has direction ratios (2, 3, 4).

#### Foot of the perpendicular
A general point of L is Q(2t, 2 + 3t, 3 + 4t). Then

PQ = (2t − 3, 3t + 3, 4t − 8).

PQ is perpendicular to the direction (2, 3, 4):

```text
2(2t − 3) + 3(3t + 3) + 4(4t − 8) = 0
4t − 6 + 9t + 9 + 16t − 32 = 0
29t − 29 = 0  →  t = 1
```

So the foot is Q(2, 5, 7).

#### Length of the perpendicular
```text
PQ = √[(2 − 3)² + (5 + 1)² + (7 − 11)²]
   = √(1 + 36 + 16)
   = √53 ≈ 7.28
```

#### Equation of the perpendicular
It is the line through P(3, −1, 11) and Q(2, 5, 7). Direction ratios: (2 − 3, 5 + 1, 7 − 11) = (−1, 6, −4).

```text
(x − 3)/(−1) = (y + 1)/6 = (z − 11)/(−4)
```

Check: dot product of (−1, 6, −4) with (2, 3, 4) is −2 + 18 − 16 = 0, so it is perpendicular to the line.

**Answer: length = √53 units; equation: (x − 3)/(−1) = (y + 1)/6 = (z − 11)/(−4).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which point represents a general point of x/2 = (y-2)/3 = (z-3)/4?
   A. Q(2t, 3t - 2, 4t - 3)
   B. Q(2t, 2 + 3t, 3 + 4t)
   C. Q(t, 2 + t, 3 + t)
   D. Q(2t + 2, 3t + 3, 4t + 4)
   Answer: B. Setting the common ratio equal to t gives x = 2t, y = 2 + 3t, z = 3 + 4t. (Key point: General point of the line)

2. How is the foot of the perpendicular found?
   A. PQ is perpendicular to the direction (2, 3, 4), so PQ . (2, 3, 4) = 0
   B. PQ is parallel to (2, 3, 4)
   C. PQ has length 1
   D. Q lies on the plane through P with normal (2, 3, 4)
   Answer: A. The dot product being zero gives 29t - 29 = 0, so t = 1 and the foot is (2, 5, 7). (Key point: Condition for foot)

3. What is the length of the perpendicular?
   A. sqrt(29)
   B. sqrt(37)
   C. sqrt(53)
   D. 53
   Answer: C. PQ = sqrt(1 + 36 + 16) = sqrt(53). (Key point: Length)

4. What are the direction ratios of the perpendicular line?
   A. (-1, 6, -4)
   B. (2, 3, 4)
   C. (1, -6, 4)
   D. (5, 4, 18)
   Answer: A. It joins P(3, -1, 11) to Q(2, 5, 7), giving (-1, 6, -4), and its dot product with (2, 3, 4) is 0. (Key point: Equation of perpendicular)

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