Show that area of the parallelogram whose diagonals are given by $\vec{a}$ and $\vec{b}$ is $\dfrac{|\vec{a}\times\vec{b}|}{2}$. Also find the area of the parallelogram whose diagonals are $2\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+3\hat{j}-\hat{k}$.

Show that area of the parallelogram whose diagonals are given by veca and vecb is dfrac|veca×vecb|2. Also find

NCERT Exemplar · Q17

\vec{a}

Karnataka PUCLong· 3mImportance★★★★★

Appeared in past exams:CBSE 2025· Set 65/4/1· 5mexact

78% · 120/153 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Parallelogram Diagonal Vectors

Parallelogram Diagonal Vectors – From Scratch

A parallelogram is a slanted rectangle: opposite sides are equal and parallel. Draw both diagonals — each runs from one corner to the opposite corner. The question is: how do we describe these diagonals using the two side vectors that start from the same corner?

The Setup

Take a parallelogram with vertices $A$ , $B$ , $C$ , $D$ in order, with $A$ at the origin. From $A$ , two vectors emerge:

$\vec{a}$ goes from $A$ to $B$ (one side)
$\vec{b}$ goes from $A$ to $D$ (the other side)

Because opposite sides are equal, $B$ to $C$ is also $\vec{b}$ and $D$ to $C$ is also $\vec{a}$ , so the fourth vertex $C$ sits at $\vec{a} + \vec{b}$ . The two diagonals run from $A$ to $C$ and from $B$ to $D$ .

The Diagonal from the Common Vertex

From $A$ to $C$ you go $\vec{a}$ then $\vec{b}$ , ending at the opposite corner:

\vec{d}_1 = \vec{a} + \vec{b}

That's the vector sum of the two sides — walk along one side then the other and you land on the opposite corner.

The Other Diagonal

$B$ is at $\vec{a}$ and $D$ is at $\vec{b}$ . To go from $B$ to $D$ , you travel from $\vec{a}$ to $\vec{b}$ :

\vec{d}_2 = \vec{b} - \vec{a}

The reverse, from $D$ to $B$ , is $\vec{a} - \vec{b}$ . Both are correct; they just differ in direction.

Note

The two diagonals are not the same length in general. They are equal only in a rectangle. The sum and difference of the side vectors give the two diagonals.

The Precise Statement

\vec{a}

Why This Matters

Vector addition — the diagonal from the common vertex is the sum of the sides. This is the parallelogram law of vector addition.
Finding midpoints — the diagonals bisect each other; both midpoints are the same point, $\frac{\vec{a} + \vec{b}}{2}$ .
Physics — the resultant of two forces acting at a point is the diagonal of the parallelogram formed by the force vectors.

A Quick Check

Take $\vec{a} = (3, 0)$ (horizontal) and $\vec{b} = (1, 2)$ (slanted). Then:

Diagonal from the common vertex: $(3,0) + (1,2) = (4,2)$ …

Start your 14-day free trial to unlock the full solution →

Sides are half the sum/difference of the diagonals, giving $\vec{a}\times\vec{b}=-2(\vec{p}\times\vec{q})$ , so area $=\frac{1}{2}|\vec{a}\times\vec{b}|$ . For the given diagonals $|\vec{a}\times\vec{b}|=\sqrt{62}$ , so the area is $\dfrac{\sqrt{62}}{2}$ square units.

Idea

The diagonals of a parallelogram are the sum and difference of its two adjacent sides. Calling the sides $\vec{p}$ and $\vec{q}$ , the diagonals are $\vec{p}+\vec{q}$ and $\vec{p}-\vec{q}$ . Since the area is $|\vec{p}\times\vec{q}|$ , we just express that cross product through the diagonals.

Proving the formula

Let $\vec{a}=\vec{p}+\vec{q}$ and $\vec{b}=\vec{p}-\vec{q}$ . Expand distributively:

\vec{a}\times\vec{b}=\vec{p}\times\vec{p}-\vec{p}\times\vec{q}+\vec{q}\times\vec{p}-\vec{q}\times\vec{q}.

Now $\vec{p}\times\vec{p}=\vec{0}$ , $\vec{q}\times\vec{q}=\vec{0}$ , and $\vec{q}\times\vec{p}=-\vec{p}\times\vec{q}$ , so

\vec{a}\times\vec{b}=-\vec{p}\times\vec{q}-\vec{p}\times\vec{q}=-2(\vec{p}\times\vec{q}).

Taking magnitudes,

|\vec{a}\times\vec{b}|=2\,|\vec{p}\times\vec{q}|\quad\Rightarrow\quad \text{Area}=|\vec{p}\times\vec{q}|=\frac{|\vec{a}\times\vec{b}|}{2}.

Applying to the given diagonals

$\vec{a}=2\hat{i}-\hat{j}+\hat{k},\ \vec{b}=\hat{i}+3\hat{j}-\hat{k}$ . …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2024Set 2024-E1 markMCQ
$\text { If } \hat{\imath}+\hat{\jmath}-\hat{k} \quad \&~ 2 \hat{\imath}-3 \hat{\jmath}+\hat{k} \text { are adjacent sides of a parallelogram, then length of its diagonals are }$
›Reveal solutionSolution
The diagonals of a parallelogram are the sum and difference of its adjacent side vectors. Using the given vectors, the diagonal lengths are $\sqrt{21}$ and $\sqrt{13}$ , so the correct option is (D).

The key idea is that in a parallelogram, if two adjacent sides are represented by vectors $\vec{a}$ and $\vec{b}$ , then the two diagonals are $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ (or $\vec{b} - \vec{a}$ , which has the same length). So we just need to compute these vector sums and differences, then find their magnitudes.

Let’s work through it step by step.

Identify the vectors

Let $\vec{a} = \hat{\imath} + \hat{\jmath} - \hat{k}$ and $\vec{b} = 2\hat{\imath} - 3\hat{\jmath} + \hat{k}$ .

Find the diagonal vectors

First diagonal: $\vec{d}_1 = \vec{a} + \vec{b}$

$\vec{d}_1 = (1+2)\hat{\imath} + (1-3)\hat{\jmath} + (-1+1)\hat{k} = 3\hat{\imath} - 2\hat{\jmath} + 0\hat{k}$

Second diagonal: $\vec{d}_2 = \vec{a} - \vec{b}$

$\vec{d}_2 = (1-2)\hat{\imath} + (1+3)\hat{\jmath} + (-1-1)\hat{k} = -1\hat{\imath} + 4\hat{\jmath} - 2\hat{k}$

Compute the lengths (magnitudes)

Length of $\vec{d}_1$ :

$|\vec{d}_1| = \sqrt{3^2 + (-2)^2 + 0^2} = \sqrt{9 + 4} = \sqrt{13}$

Length of $\vec{d}_2$ :
Start your 14-day free trial to see the full solution →
COMEDK 2025Set 2025-A1 markMCQ
$|\vec{a}|=2 \sqrt{2}$
›Reveal solutionSolution

Diagonals of the parallelogram are $\vec p+\vec q=3\vec a-2\vec b$ and $\vec p-\vec q=\vec a-4\vec b$ ; their lengths multiply to $12\sqrt{26}$ .

Given: $|\vec a|=2\sqrt2\Rightarrow|\vec a|^2=8$ , $|\vec b|=3\Rightarrow|\vec b|^2=9$ , angle $=\dfrac{\pi}{4}$ .
$\vec a\cdot\vec b=|\vec a||\vec b|\cos\tfrac{\pi}{4}=2\sqrt2\cdot 3\cdot\tfrac{1}{\sqrt2}=6.$
The diagonals of a parallelogram with adjacent sides $\vec p,\vec q$ are $\vec p+\vec q$ and $\vec p-\vec q$ , where $\vec p=2\vec a-3\vec b$ and $\vec q=\vec a+\vec b$ .

First diagonal:
$\vec p+\vec q=3\vec a-2\vec b,\qquad |\vec p+\vec q|^2=9|\vec a|^2-12(\vec a\cdot\vec b)+4|\vec b|^2=72-72+36=36,$
so $|\vec p+\vec q|=6$ . …
Start your 14-day free trial to see the full solution →

←Previous questionQ16 — If veca, vecb, vecc determine the vertices of a triangle, show that (1…→Next questionQ18 — If veca=hati+hatj+hatk and vecb=hatj-hatk, find a vector vecc such tha…

Q.Show that area of the parallelogram whose diagonals are given by $\vec{a}$ and $\vec{b}$ is $\dfrac{|\vec{a}\times\vec{b}|}{2}$ . Also find the area of the parallelogram whose diagonals are $2\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+3\hat{j}-\hat{k}$ .

Concept understanding — Parallelogram Diagonal Vectors

Parallelogram Diagonal Vectors – From Scratch

The Setup

The Diagonal from the Common Vertex

The Other Diagonal

The Precise Statement

Why This Matters

A Quick Check

Idea

Proving the formula

Applying to the given diagonals

Method: Area of a Parallelogram from Its Diagonals

Steps

Common Mistakes

Idea

Proving the formula

Applying to the given diagonals

🎓Unlock everything free for 14 days

Unlock everything free for 14 days