NIMCET 2009 — Mathematics PYQ

Recall the property $|\mathbf{x}-\mathbf{y}{|}^2=|\mathbf{x}{|}^2+|\mathbf{y}{|}^2-2(\mathbf{x}\cdot \mathbf{y})$. Since $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ are unit vectors, we have $|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=1$.

Expanding each term:

• $|\mathbf{a}-\mathbf{b}{|}^2=|\mathbf{a}{|}^2+|\mathbf{b}{|}^2-2(\mathbf{a}\cdot \mathbf{b})=1+1-2(\mathbf{a}\cdot \mathbf{b})=2-2(\mathbf{a}\cdot \mathbf{b})$

• $|\mathbf{b}-\mathbf{c}{|}^2=|\mathbf{b}{|}^2+|\mathbf{c}{|}^2-2(\mathbf{b}\cdot \mathbf{c})=1+1-2(\mathbf{b}\cdot \mathbf{c})=2-2(\mathbf{b}\cdot \mathbf{c})$

• $|\mathbf{c}-\mathbf{a}{|}^2=|\mathbf{c}{|}^2+|\mathbf{a}{|}^2-2(\mathbf{c}\cdot \mathbf{a})=1+1-2(\mathbf{c}\cdot \mathbf{a})=2-2(\mathbf{c}\cdot \mathbf{a})$

2. Sum the Expressions

Let the total sum be $S$:

3. Use the Property of Vector Sums

We know that the square of the magnitude of any vector is always non-negative:

Expanding this:

Multiply by $-1$ (this reverses the inequality):

4. Final Calculation

Substitute this inequality back into our expression for $S$:

Conclusion:

The expression does not exceed $9$.

Final Answer:

The correct option is (a) 9.