Question 141 markMCQSection A

The value of m for which the points with position vectors −i^−j^+2k^-\hat{i} - \hat{j} + 2\hat{k}, 2i^+mj^+5k^2\hat{i} + m\hat{j} + 5\hat{k} and 3i^+11j^+6k^3\hat{i} + 11\hat{j} + 6\hat{k} are collinear, is

Vector Algebra
Collinearity of Points using Position Vectors

Options

(A)8
(B)-8
(C)2
(D)52\frac{5}{2}

Official answer

Correct option: (A) 8


Let A=(−1,−1,2),B=(2,m,5),C=(3,11,6)A=(-1,-1,2), B=(2,m,5), C=(3,11,6). Vector AB=3i+(m+1)j+3kAB = 3i+(m+1)j+3k and AC=4i+12j+4kAC = 4i+12j+4k. For collinearity, AB must be a scalar multiple of AC. Comparing the i-components: 3=4t3=4t gives t=34t=\frac{3}{4}; matching j-components: m+1=12×34=9m+1 = 12 \times \frac{3}{4} = 9, so m=8m=8.

Answer keywords:collinear pointsposition vectorsvector algebraparallel vectorsscalar multipleproportional components

Marking scheme

  1. 1.1 mark: awarded only for selecting option (A) 8; no partial credit.

Hint

Find AB and AC as vectors, then equate the ratio of their corresponding (i,j,k)(i, j, k) components since collinear points give parallel connecting vectors.

Quick oral answer

AB and AC must be parallel for collinear points; comparing components (3,m+1,3)(3,m+1,3) proportional to (4,12,4)(4,12,4) gives t=34t=\frac{3}{4} and m=8m=8.

Common mistakes

  1. 1.Sign errors when subtracting position vectors with negative components (e.g., m−(−1)m-(-1)).
  2. 2.Using inconsistent component ratios to find t, leading to an incorrect value of m.
  3. 3.Confusing the collinearity condition (vectors parallel) with perpendicularity (dot product zero).

Analysis and explanation

Concept: Three points are collinear if and only if the vectors joining any two pairs of them are parallel (scalar multiples of each other).


With A=(−1,−1,2),B=(2,m,5),C=(3,11,6)A=(-1,-1,2), B=(2,m,5), C=(3,11,6):


  • AB=(2−(−1))i+(m−(−1))j+(5−2)k=3i+(m+1)j+3kAB = (2-(-1))i + (m-(-1))j + (5-2)k = 3i + (m+1)j + 3k
  • AC=(3−(−1))i+(11−(−1))j+(6−2)k=4i+12j+4kAC = (3-(-1))i + (11-(-1))j + (6-2)k = 4i + 12j + 4k

For A, B, C to be collinear, AB=t⋅ACAB = t \cdot AC for some scalar t. Comparing i and k components (both give 3=4t3=4t), t=34t = \frac{3}{4}. Substituting into the j-component: m+1=12t=9m+1 = 12t = 9, so m=8m = 8.


Why the distractors fail:

  • (B) -8 is a sign error, likely from mishandling the subtraction m-(-1) or an error transposing across the equation m+1=9.
  • (C) 2 and (D) 5/2 result from using the wrong ratio (e.g., matching only one component incorrectly or computing t from mismatched pairs of components), giving an inconsistent, incorrect value of m.

Exam tip: Always verify collinearity using at least two independent component ratios (e.g., i and k here) to confirm consistency before solving for the unknown using the third.

Interesting facts

The collinearity test using proportional vector components is essentially the 3D extension of the 2D condition that three points on a line have equal slopes.

This same collinearity technique underlies collision and alignment detection algorithms used in 3D computer graphics and robotics.

Question details

Q14 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Vector Algebra
Topic
Collinearity of Points using Position Vectors
Bloom's level
apply
Syllabus topic
Unit 4: Vector Algebra — Collinearity of Points using Vectors
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

What condition must hold for three points given by position vectors to be collinear?

The vector joining any two of the points must be a scalar multiple of the vector joining another pair of the same points — i.e., AB=t⋅ACAB = t \cdot AC for some real number t.

Why do we check two components before solving for the unknown?

Checking two independent components (here, i and k) confirms the value of the scalar t consistently, ensuring the collinearity condition truly holds before using it to solve for the unknown m in the third component.