Sample 2025-26 Class 12 Maths Q25 Solution: Area of the parallelogram on OA and the median OP of ΔOAB

The question

The two vectors $\hat i + \hat j + \hat k$ and $3\hat i - \hat j + 3\hat k$ represent the two sides $OA$ and $OB$ ⁠, respectively of a $\Delta OAB$ ⁠, where $O$ is the origin. The point $P$ lies on $AB$ such that $OP$ is a median. Find the area of the parallelogram formed by the two adjacent sides as $OA$ and $OP$ ⁠.

Exam answer

As $OP$ is a median, $P$ is the mid-point of $AB$ ⁠. So

\begin{aligned} \overrightarrow{OP} &= \frac12\left(\overrightarrow{OA} + \overrightarrow{OB}\right) \\ &= \frac12(4\hat i + 4\hat k) = 2\hat i + 2\hat k \end{aligned}

Why?

A median joins a vertex to the mid-point of the opposite side. The position vector of the mid-point of $AB$ is the average of the position vectors of $A$ and $B$ ⁠:

\overrightarrow{OP} = \frac{\vec a + \vec b}{2}

Add the components: $1 + 3 = 4$ ⁠, $1 - 1 = 0$ and $1 + 3 = 4$ ⁠, so $\overrightarrow{OA} + \overrightarrow{OB} = 4\hat i + 0\hat j + 4\hat k$ ⁠.

\begin{aligned} \frac12(4\hat i + 4\hat k) &= \frac42\hat i + \frac42\hat k \\ &= 2\hat i + 2\hat k \end{aligned}

Area of the parallelogram with adjacent sides OA and OP:

\begin{aligned} &= \left|\overrightarrow{OA} \times \overrightarrow{OP}\right| \\ &= \left|\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 1 \\ 2 & 0 & 2 \end{vmatrix}\right| \end{aligned}

Why?

The area of a parallelogram with adjacent sides $\vec a$ and $\vec b$ is the magnitude of their cross product:

\text{Area} = |\vec a \times \vec b|

The cross product is the determinant with $\hat i, \hat j, \hat k$ in the first row, the components of $\overrightarrow{OA}$ (1, 1, 1) in the second and those of $\overrightarrow{OP}$ (2, 0, 2) in the third.

\begin{aligned} \text{Area} &= \big|\hat i(2 - 0) - \hat j(2 - 2) \\ &\qquad + \hat k(0 - 2)\big| \\ &= \lvert 2\hat i - 0\hat j - 2\hat k\rvert \\ &= \sqrt{4 + 0 + 4} \end{aligned}

Why?

Expand along the first row, with the signs $+, -, +$ ⁠. Each entry multiplies the 2 by 2 determinant left when its row and column are covered:

\begin{aligned} \hat i&: \begin{vmatrix} 1 & 1 \\ 0 & 2 \end{vmatrix} \\ &= (1)(2) - (1)(0) \\ &= 2 - 0 = 2 \\[4pt] \hat j&: \begin{vmatrix} 1 & 1 \\ 2 & 2 \end{vmatrix} \\ &= (1)(2) - (1)(2) \\ &= 2 - 2 = 0 \\[4pt] \hat k&: \begin{vmatrix} 1 & 1 \\ 2 & 0 \end{vmatrix} \\ &= (1)(0) - (1)(2) \\ &= 0 - 2 = -2 \end{aligned}

With the signs: $+2\hat i - (0)\hat j + (-2)\hat k = 2\hat i - 2\hat k$ ⁠.

The magnitude is $|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}$ ⁠, so we get $\sqrt8$ ⁠. Then $\sqrt8 = \sqrt{4 \times 2} = 2\sqrt2$ ⁠.

\begin{aligned} &\sqrt{(2)^2 + (0)^2 + (-2)^2} \\ &= \sqrt{4 + 0 + 4} = \sqrt8 \\ &= \sqrt{4 \times 2} = \sqrt4 \times \sqrt2 \\ &= 2\sqrt2 \end{aligned}

$\therefore$ Area $= 2\sqrt2$ square units.

Where this comes from

Chapter 10, Vector Algebra: Vector product: area of a parallelogram.

Question type: Section formula + Cross product, and the area of a parallelogram (combines 2 types)

How to spot it: click to see

Type 3. Section formula: the point that divides PQ

Spot it: "divides the line joining P and Q in the ratio 2 : 1", internally or externally.

Here: First, the median OP goes to the midpoint of AB, so $\overrightarrow{OP} = \frac12(\overrightarrow{OA} + \overrightarrow{OB}) = 2\hat i + 2\hat k$ ⁠.

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

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

Here: Then OA and OP are adjacent sides, so the area is $|\overrightarrow{OA} \times \overrightarrow{OP}| = |2\hat i - 2\hat k|$ ⁠.

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Q25: Area of the parallelogram on OAOAOA and the median OPOPOP of ΔOAB\Delta OABΔOAB

Q25: Area of the parallelogram on $OA$ and the median $OP$ of $\Delta OAB$