---
title: "Lines intersect: point and plane through them"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:lines-intersect-point-plane-through"
source: https://www.bctengineer.com/q/semester_2/mathematics/lines-intersect-point-plane-through
site: BCT Engineering
---

# Lines intersect: point and plane through them

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2067 Chaitra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/lines-intersect-point-plane-through
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Prove that the lines (x−1)/2 = (y+1)/(−3) = (z+10)/8 and (x−4)/1 = (y+3)/(−4) = (z+1)/7 intersect. Also find their point of intersection and plane through them.

## Short answer

Equating parametric forms gives t=2, s=1, intersection point (5,-7,6). The plane containing both lines has normal (11,-6,-5) and equation 11x - 6y - 5z = 67.

## Answer

Two lines intersect if a common point exists; then they are coplanar.

Line 1: (x−1)/2 = (y+1)/(−3) = (z+10)/8 = t, so (1 + 2t, −1 − 3t, −10 + 8t).
Line 2: (x−4)/1 = (y+3)/(−4) = (z+1)/7 = s, so (4 + s, −3 − 4s, −1 + 7s).

#### Intersection
Equate the coordinates:

```text
1 + 2t = 4 + s          → 2t − s = 3     ... (i)
−1 − 3t = −3 − 4s       → −3t + 4s = −2  ... (ii)
−10 + 8t = −1 + 7s      → 8t − 7s = 9    ... (iii)
```
From (i), s = 2t − 3. Put in (ii): −3t + 8t − 12 = −2, so t = 2 and s = 1.

Check (iii): 8(2) − 7(1) = 9 ✓. All three equations are satisfied, so the lines intersect.

**Point of intersection:** t = 2 gives (5, −7, 6). (s = 1 gives (5, −7, 6) as well.)

#### Plane through them
Normal = d₁ × d₂ with d₁ = (2, −3, 8), d₂ = (1, −4, 7):

```text
n = ((−3)(7) − 8(−4), 8(1) − 2(7), 2(−4) − (−3)(1))
  = (−21 + 32, 8 − 14, −8 + 3) = (11, −6, −5)
```
Plane through (5, −7, 6):

11(x − 5) − 6(y + 7) − 5(z − 6) = 0 → 11x − 6y − 5z − 55 − 42 + 30 = 0.

**Answer: the lines meet at (5, −7, 6) and lie in the plane 11x − 6y − 5z = 67.**

Check: the point (1, −1, −10) of line 1: 11 + 6 + 50 = 67 ✓.

## Check yourself

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

1. How is it shown the lines intersect?
   A. Check d1 . d2 = 0
   B. Equate coordinates with parameters t and s and check all three equations hold
   C. Check the lines are parallel
   D. Check d1 = d2
   Answer: B. A common point exists only if one (t, s) satisfies all three. (Key point: Intersection method)

2. Solving the first two equations gives:
   A. t = 1, s = 2
   B. t = 2, s = -1
   C. t = 2, s = 1
   D. t = 3, s = 3
   Answer: C. s = 2t - 3 into -3t + 4s = -2 gives t = 2. (Key point: Parameters)

3. The point of intersection is:
   A. (5, -7, 6)
   B. (1, -1, -10)
   C. (5, 7, 6)
   D. (4, -3, -1)
   Answer: A. t = 2: (1+4, -1-6, -10+16). (Key point: Point)

4. The plane through them is:
   A. 11x - 6y - 5z = -67
   B. 11x - 6y - 5z = 67
   C. 11x + 6y - 5z = 67
   D. x - y - z = 67
   Answer: B. Normal d1 x d2 = (11, -6, -5) through (5, -7, 6). (Key point: Plane)

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