CBSE Class 12 Mathematics sample paper 2026-27: question 23(B)

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Step-by-step solution

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1. Let $\overrightarrow{AB}=\vec{a}$, $\overrightarrow{BC}=\vec{b}$, $\overrightarrow{CD}=\vec{c}$, $\overrightarrow{DA}=\vec{d}$. In a rectangle, $\vec{c}=-\vec{a}$ and $\vec{d}=-\vec{b}$. Then $\vec{u}=\overrightarrow{AC}=\vec{a}+\vec{b}$ …(i) and $\vec{v}=\overrightarrow{BD}=\vec{b}+\vec{c}=\vec{b}-\vec{a}$ …(ii).

   ½ mark

2. Adding (i) and (ii): $\vec{b}=\frac{1}{2}(\vec{u}+\vec{v})$. Subtracting (ii) from (i): $\vec{a}=\frac{1}{2}(\vec{u}-\vec{v})$.

   1 mark

3. So $\overrightarrow{AB}=\frac{1}{2}(\vec{u}-\vec{v})$, $\overrightarrow{BC}=\frac{1}{2}(\vec{u}+\vec{v})$, $\overrightarrow{CD}=\frac{1}{2}(\vec{v}-\vec{u})$, $\overrightarrow{DA}=-\frac{1}{2}(\vec{u}+\vec{v})$.

   ½ mark

Answer: $\overrightarrow{AB}=\frac{1}{2}(\vec{u}-\vec{v})$, $\overrightarrow{BC}=\frac{1}{2}(\vec{u}+\vec{v})$, $\overrightarrow{CD}=\frac{1}{2}(\vec{v}-\vec{u})$, $\overrightarrow{DA}=-\frac{1}{2}(\vec{u}+\vec{v})$

Revise the chapter

• Chapter: Vector Algebra (notes, formulas, where marks are lost and every past question)