Vectors Ques 6
- If $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are unit vectors, then $|\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}|^{2}+|\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}|^{2}+|\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}}|^{2}$ does not exceed
(2001, 2M)
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Show Answer
Answer:
Correct Answer: 6.(b)
Solution:
- Now, $(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}})^{2}=\Sigma \overrightarrow{\mathbf{a}}^{2}+2 \sum_{i<j} \overrightarrow{\mathbf{a}}{i} \cdot \overrightarrow{\mathbf{a}}{j} \geq 0$
$\Rightarrow \quad 2 \Sigma \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} \geq-3$
$[\because|\overrightarrow{\mathbf{a}}|=|\overrightarrow{\mathbf{b}}|=|\overrightarrow{\mathbf{c}}|=1]$
Now, $\quad \Sigma|\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}|^{2}=2 \Sigma \overrightarrow{\mathbf{a}}^{2}-2 \Sigma \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} \leq 2$ (3) $+3$
BETA
