Aspire Faculty ID #16424 · Topic: NIMCET 2009 · Just now
NIMCET 2009

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, then $|\vec{a}-\vec{b}|^2 + |\vec{b}-\vec{c}|^2 + |\vec{c}-\vec{a}|^2$ does not exceed:

Solution

Expand

\[|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b} = 2 - 2\vec{a}\cdot\vec{b}\]

\[|\vec{b}-\vec{c}|^2 = 2 - 2\vec{b}\cdot\vec{c}\]

\[|\vec{c}-\vec{a}|^2 = 2 - 2\vec{c}\cdot\vec{a}\]

Add:

\[= 6 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})\]

Find Maximum :

\[|\vec{a} + \vec{b} + \vec{c}|^2 \geq 0\]

\[3 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \geq 0\]

\[\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} \geq -\frac{3}{2}\]

Final Answer:

\[6 - 2 \times \left(-\frac{3}{2}\right) = 6 + 3 = \boxed{9}\]

✅ Expression does not exceed 9

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