Question 242 marksVery Short AnswerSection B

Vectors a=3i−2j+2ka = 3i - 2j + 2k and b=i+2kb = i + 2k represent the two adjacent sides of a parallelogram. Find the vectors representing its diagonals and hence find their lengths.

Vector Algebra
Vectors - Parallelogram Diagonals

Official answer

The diagonals are d1=a+b=4i−2j+4kd_1 = a + b = 4i - 2j + 4k with length 6, and d2=a−b=2i−2jd_2 = a - b = 2i - 2j with length 222\sqrt{2}.


Working: For a parallelogram with adjacent sides a=3i−2j+2ka = 3i-2j+2k and b=i+2kb = i+2k, the diagonals are a+b and a−b. a+b=4i−2j+4ka+b = 4i-2j+4k, ∣a+b∣=16+4+16=36=6|a+b| = \sqrt{16+4+16} = \sqrt{36} = 6. a−b=2i−2j+0ka-b = 2i-2j+0k, ∣a−b∣=4+4=8=22|a-b| = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}.

Answer keywords:parallelogram law of vectorsdiagonal vectorsmagnitude of a vectorvector additionvector subtractionadjacent sides

Marking scheme

  1. 1.1 mark: correctly finding both diagonal vectors a+b=4i−2j+4ka+b = 4i-2j+4k and a−b=2i−2ja-b = 2i-2j.
  2. 2.1 mark: correctly computing both magnitudes, 6 and 222\sqrt{2}.

Hint

The diagonals of a parallelogram with adjacent sides a and b are a+ba+b and a−ba-b; find each and then their magnitudes.

Quick oral answer

For adjacent sides a and b of a parallelogram, the diagonals are a+ba+b and a−ba-b; computing gives diagonals 4i−2j+4k4i-2j+4k (length 6) and 2i−2j2i-2j (length 222\sqrt{2}).

Common mistakes

  1. 1.Finding only one diagonal (usually a+ba+b) and forgetting the second (a−ba-b).
  2. 2.Sign error when subtracting components, inconsistently mixing a−ba-b and b−ab-a in the final answer.
  3. 3.Arithmetic mistakes in computing 36\sqrt{36} or 8\sqrt{8}, e.g. leaving 8\sqrt{8} unsimplified instead of writing 222\sqrt{2}.

Analysis and explanation

This question uses the standard parallelogram law: if two adjacent sides are represented by vectors a and b, the two diagonals are represented by a+ba+b and a−ba-b.


Concept: The diagonal from the common starting vertex to the opposite vertex is a+ba+b (vector addition, tip-to-tail), while the diagonal connecting the tips of a and b is a−ba-b (or b−ab-a, same length, opposite direction).


Key steps: Add and subtract the given vectors component-wise, then apply the magnitude formula x2+y2+z2\sqrt{x^2+y^2+z^2} to each resulting diagonal vector.


Exam trap: A common error is computing only one diagonal, or mixing up which combination (a+ba+b vs a−ba-b) represents which diagonal — both must be found and both lengths reported since the question explicitly asks for the diagonals and their lengths.


Real-world link: This vector approach to parallelogram diagonals underlies how engineers compute resultant and relative displacement/force vectors in truss and structural analysis, where two members meeting at a joint are modelled exactly like adjacent sides of a parallelogram.

Interesting facts

This is a direct application of the Parallelogram Law of Vector Addition, one of the oldest results in vector geometry, tracing back to work by Simon Stevin in the late 16th century on the composition of forces.

Whether a parallelogram is a rhombus can be tested using exactly this diagonal method: a rhombus's diagonals are always perpendicular, i.e. (a+b)⋅(a−b)=0(a+b)\cdot(a-b) = 0, which simplifies to ∣a∣=∣b∣|a| = |b|.

Question details

Q24 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section B
Marks
2
Type
Very Short Answer
Chapter
Vector Algebra
Topic
Vectors - Parallelogram Diagonals
Bloom's level
apply
Syllabus topic
Unit 4: Vector Algebra — Vector Addition/Subtraction, Magnitude of a Vector, Geometric Application
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

What is the formula for the diagonals of a parallelogram in vector form?

If a and b represent two adjacent sides of a parallelogram, its diagonals are represented by the vectors a+ba+b and a−ba-b.

Why are two diagonals computed instead of one?

A parallelogram has exactly two diagonals, and the question explicitly asks for both the diagonal vectors and their lengths, so both a+ba+b and a−ba-b (and their magnitudes) must be reported for full marks.