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Question

Mathematics Question on Vector Algebra

If a→,b→\overrightarrow{a}, \overrightarrow{b} and c→ \overrightarrow{c} are unit vectors, then ∣a→−b→∣2+∣b→−c→∣2+∣c→−a→∣2|\overrightarrow{a}-\overrightarrow{b} |^2+| \overrightarrow{b}-\overrightarrow{c}|^2+| \overrightarrow{c}-\overrightarrow{a}|^2 does not exceed

A

4

B

9

C

8

D

6

Answer

9

Explanation

Solution

Now, (a→+b→+c→)2=Σa→2+2Σa→.b→≥0(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})^2=\boldsymbol{\Sigma}\overrightarrow{a}^2+2\boldsymbol{\Sigma}\overrightarrow{a}.\overrightarrow{b}\ge 0
\Rightarrow \hspace25mm 2\boldsymbol{\Sigma}\overrightarrow{a}.\overrightarrow{b} \ge -3 [\because |\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1]
Now, Σ∣a→−b→∣2=2Σa→2−2Σa→.b→≥2(3)+3=9\boldsymbol{\Sigma}|\overrightarrow{a}-\overrightarrow{b}|^2=2\boldsymbol{\Sigma}\overrightarrow{a}^2-2\boldsymbol{\Sigma}\overrightarrow{a}.\overrightarrow{b}\ge 2(3)+3=9