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

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

The diagonals are **$d_1 = a + b = 4i - 2j + 4k$** with length **6**, and **$d_2 = a - b = 2i - 2j$** with length **$2\sqrt{2}$**. **Working:** For a parallelogram with adjacent sides $a = 3i-2j+2k$ and $b = i+2k$, the diagonals are a+b and a−b. $a+b = 4i-2j+4k$, $|a+b| = \sqrt{16+4+16} = \sqrt{36} = 6$. $a-b = 2i-2j+0k$, $|a-b| = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$.

Vector Algebra

Vectors - Parallelogram Diagonals

Official Answer

The diagonals are $d_1 = a + b = 4i - 2j + 4k$ with length 6, and $d_2 = a - b = 2i - 2j$ with length $2\sqrt{2}$ .

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

parallelogram law of vectorsdiagonal vectorsmagnitude of a vectorvector additionvector subtractionadjacent sides

Marking Scheme

11 mark: correctly finding both diagonal vectors $a+b = 4i-2j+4k$ and $a-b = 2i-2j$ .
21 mark: correctly computing both magnitudes, 6 and $2\sqrt{2}$ .

Hint

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

Quick Oral Answer

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

Analysis & 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 $a+b$ and $a-b$ .

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

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

Exam trap: A common error is computing only one diagonal, or mixing up which combination ( $a+b$ vs $a-b$ ) 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.

Common Mistakes

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

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. $(a+b)\cdot(a-b) = 0$ , which simplifies to $|a| = |b|$ .

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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 $a+b$ and $a-b$ .

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 $a+b$ and $a-b$ (and their magnitudes) must be reported for full marks.