CBSE Class 12 Mathematics 2026: question 30(b)

Hints

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Step-by-step solution

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1. Since the vectors are unit vectors, $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$.

   ½ mark

2. Expand each square using $|\vec{p}-\vec{q}{|}^2=|\vec{p}{|}^2+|\vec{q}{|}^2-2\vec{p}\cdot \vec{q}$:

   $|\vec{a}-\vec{b}{|}^2+|\vec{b}-\vec{c}{|}^2+|\vec{c}-\vec{a}{|}^2=2|\vec{a}{|}^2+2|\vec{b}{|}^2+2|\vec{c}{|}^2-2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})$

   1½ marks

3. Rewrite this as $3(|\vec{a}{|}^2+|\vec{b}{|}^2+|\vec{c}{|}^2)-\left(|\vec{a}{|}^2+|\vec{b}{|}^2+|\vec{c}{|}^2+2\vec{a}\cdot \vec{b}+2\vec{b}\cdot \vec{c}+2\vec{c}\cdot \vec{a}\right)=9-|\vec{a}+\vec{b}+\vec{c}{|}^2$

   ½ mark

4. Since $|\vec{a}+\vec{b}+\vec{c}{|}^2\ge 0$, it follows that $|\vec{a}-\vec{b}{|}^2+|\vec{b}-\vec{c}{|}^2+|\vec{c}-\vec{a}{|}^2\le 9$. Hence proved.

   ½ mark

Answer: LHS $=9-|\vec{a}+\vec{b}+\vec{c}{|}^2\le 9$ (proved)

Revise the chapter

• Chapter: Vector Algebra (notes, formulas, where marks are lost and every past question)
• Topic: Scalar (Dot) Product of Vectors