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}

Assam AhsecLong· 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.

AHSEC Higher Secondary (HS) Final Examination 2019Set ANNUAL6 marks
$(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 2(\vec{a}\times\vec{b})$
›Reveal solutionSolution

Main: expand the cross product using bilinearity and $\vec a\times\vec a=\vec b\times\vec b=\vec 0$ , $\vec b\times\vec a=-\vec a\times\vec b$ ; then use "area $=\tfrac12|\vec d_1\times\vec d_2|$ " for diagonals $\vec d_1,\vec d_2$ . OR: the required direction is perpendicular to both plane normals, i.e. their cross product.

Main question — identity.

$(\vec a-\vec b)\times(\vec a+\vec b) = \vec a\times\vec a + \vec a\times\vec b - \vec b\times\vec a - \vec b\times\vec b.$

$\vec a\times\vec a=\vec0$ , $\vec b\times\vec b=\vec0$ , and $\vec b\times\vec a = -\vec a\times\vec b$ . So

$(\vec a-\vec b)\times(\vec a+\vec b) = \vec0 + \vec a\times\vec b -(-\vec a\times\vec b) - \vec 0 = 2(\vec a\times\vec b).$

Application. If the diagonals of a parallelogram are $\vec d_1$ and $\vec d_2$ , then (writing $\vec d_1=\vec a+\vec b,\ \vec d_2=\vec a-\vec b$ for the two adjacent sides $\vec a,\vec b$ ) the identity just proved gives $\vec d_2\times\vec d_1 = -2(\vec a\times\vec b)$ , i.e. $|\vec a\times\vec b| = \tfrac12|\vec d_1\times\vec d_2|$ . Since the parallelogram's area is $|\vec a\times\vec b|$ , we get:

$\text{Area} = \frac12|\vec d_1\times\vec d_2|.$

Here $\vec d_1 = 3\hat i+\hat j-2\hat k$ , $\vec d_2=\hat i-3\hat j+4\hat k$ .

$\vec d_1\times\vec d_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\3&1&-2\\1&-3&4\end{vmatrix} = \hat i[1(4)-(-2)(-3)] - \hat j[3(4)-(-2)(1)] + \hat k[3(-3)-1(1)] = -2\hat i-14\hat j-10\hat k.$

$|\vec d_1\times\vec d_2| = \sqrt{(-2)^2+(-14)^2+(-10)^2} = \sqrt{4+196+100} = \sqrt{300} = 10\sqrt3.$

$\text{Area} = \frac12(10\sqrt3) = 5\sqrt3 \text{ square units}.$

…

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