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CBSE Board Examination 2025 · Section B · 2 marks

Q23: Area of the parallelogram with diagonals 2i^−j^+k^2\hat i - \hat j + \hat k and i^+3j^−k^\hat i + 3\hat j - \hat k

Board 2025Ch 10 · Vector Algebra
The question

The diagonals of a parallelogram are given by a⃗=2i^−j^+k^\vec a = 2\hat i - \hat j + \hat k and b⃗=i^+3j^−k^\vec b = \hat i + 3\hat j - \hat k⁠. Find the area of the parallelogram.

Exam answer
a⃗×b⃗=∣i^j^k^2−1113−1∣\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{vmatrix}
Why?

The cross product is found as a determinant. The first row holds i^,j^,k^\hat i, \hat j, \hat k⁠, the second row the components of a⃗\vec a⁠, and the third row those of b⃗\vec b⁠.

Here a⃗\vec a has components 2,−1,12, -1, 1 and b⃗\vec b has 1,3,−11, 3, -1⁠.

=i^(1−3)−j^(−2−1)+k^(6+1)=−2i^+3j^+7k^\begin{aligned} &= \hat i(1 - 3) - \hat j(-2 - 1) \\ &\quad + \hat k(6 + 1) \\ &= -2\hat i + 3\hat j + 7\hat k \end{aligned}
Why?

Expand along the first row. Each unit vector multiplies the 2 by 2 determinant left when its row and column are removed:

i^: ∣−113−1∣=(−1)(−1)−(1)(3)=1−3=−2j^: ∣211−1∣=(2)(−1)−(1)(1)=−2−1=−3k^: ∣2−113∣=(2)(3)−(−1)(1)=6+1=7\begin{aligned} \hat i&:\ \begin{vmatrix} -1 & 1 \\ 3 & -1 \end{vmatrix} \\ &= (-1)(-1) - (1)(3) \\ &= 1 - 3 = -2 \\[4pt] \hat j&:\ \begin{vmatrix} 2 & 1 \\ 1 & -1 \end{vmatrix} \\ &= (2)(-1) - (1)(1) \\ &= -2 - 1 = -3 \\[4pt] \hat k&:\ \begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix} \\ &= (2)(3) - (-1)(1) \\ &= 6 + 1 = 7 \end{aligned}

Now apply the signs ++⁠, −-⁠, ++⁠:

i^(−2)−j^(−3)+k^(7)=−2i^+3j^+7k^\begin{aligned} &\hat i(-2) - \hat j(-3) + \hat k(7) \\ &= -2\hat i + 3\hat j + 7\hat k \end{aligned}

Watch out: the j^\hat j term carries a minus sign in front.

The area of a parallelogram with diagonals a⃗\vec a and b⃗\vec b is 12∣a⃗×b⃗∣\frac12|\vec a \times \vec b|⁠.

Why?

Let the adjacent sides be p⃗\vec p and q⃗\vec q⁠. Then the diagonals are a⃗=p⃗+q⃗\vec a = \vec p + \vec q and b⃗=q⃗−p⃗\vec b = \vec q - \vec p⁠, and

a⃗×b⃗=(p⃗+q⃗)×(q⃗−p⃗)=2(p⃗×q⃗)\begin{aligned} \vec a \times \vec b &= (\vec p + \vec q) \times (\vec q - \vec p) \\ &= 2(\vec p \times \vec q) \end{aligned}

(using p⃗×p⃗=0⃗\vec p \times \vec p = \vec 0 and q⃗×p⃗=−p⃗×q⃗\vec q \times \vec p = -\vec p \times \vec q⁠). Term by term:

(p⃗+q⃗)×(q⃗−p⃗)=p⃗×q⃗−p⃗×p⃗+q⃗×q⃗−q⃗×p⃗=p⃗×q⃗−0⃗+0⃗+p⃗×q⃗=2(p⃗×q⃗)\begin{aligned} &(\vec p + \vec q) \times (\vec q - \vec p) \\ &= \vec p \times \vec q - \vec p \times \vec p \\ &\quad + \vec q \times \vec q - \vec q \times \vec p \\ &= \vec p \times \vec q - \vec 0 + \vec 0 + \vec p \times \vec q \\ &= 2(\vec p \times \vec q) \end{aligned}

The area is ∣p⃗×q⃗∣|\vec p \times \vec q|⁠, which is half of ∣a⃗×b⃗∣|\vec a \times \vec b|⁠.

Watch out: with adjacent sides there is no 12\frac12⁠; with diagonals there is.

Area=12(−2)2+32+72=124+9+49\begin{aligned} \text{Area} &= \frac12\sqrt{(-2)^2 + 3^2 + 7^2} \\ &= \frac12\sqrt{4 + 9 + 49} \end{aligned}
Why?

The magnitude of a vector is the square root of the sum of the squares of its components:

∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}

Here 4+9+49=624 + 9 + 49 = 62⁠.

(−2)2=4, 32=9, 72=494+9=1313+49=62Area=1262=622\begin{aligned} (-2)^2 &= 4,\ 3^2 = 9,\ 7^2 = 49 \\ 4 + 9 &= 13 \\ 13 + 49 &= 62 \\ \text{Area} &= \frac12\sqrt{62} = \frac{\sqrt{62}}{2} \end{aligned}

∴\therefore Area of the parallelogram =622= \dfrac{\sqrt{62}}{2} sq. units.

Where this comes from

Chapter 10, Vector Algebra: Area of a parallelogram from its diagonals.

Question type: Cross product, and the area of a parallelogram

How to spot it: click to see

Type 8. Cross product, and the area of a parallelogram

Spot it: "find a⃗×b⃗\vec a\times\vec b", "find ∣a⃗×b⃗∣|\vec a\times\vec b|", or "the area of the parallelogram whose adjacent sides are".

Here: The two given vectors are diagonals, not sides. Find their cross product −2i^+3j^+7k^-2\hat i + 3\hat j + 7\hat k and take half its length.

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