Let a→ , b→ and c→ are three unit vectors such that |a→−b→|2+|b→−c→|2+|c→+a→|2=3 , then which of following option(s) is/are correct?

|a→−b→|2+|b→−c→|2+|c→+a→|2=3⇒|a→+c→−b→|2=0 , a→+c→=b→⇒a→⋅c→=−12So , |a→+b→+c→|=2|b→|=2 , |a→+2b→+3c→|=|3a→+5c→|=19

Let a→ , b→ and c→ are three unit vectors such that |a→−b→|2+|b→−c→|2+|c→+a→|2=3 , then which of following option(s) is/are correct?

|a→−b→|2+|b→−c→|2+|c→+a→|2=3⇒|a→+c→−b→|2=0 , a→+c→=b→⇒a→⋅c→=−12So , |a→+b→+c→|=2|b→|=2 , |a→+2b→+3c→|=|3a→+5c→|=19

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Let $\vec{a}$ , $\vec{b}$ and $\vec{c}$ are three unit vectors such that ${| \vec{a} - \vec{b} |}^{2} + {| \vec{b} - \vec{c} |}^{2} + {| \vec{c} + \vec{a} |}^{2} = 3$ , then which of following option(s) is/are correct?

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$\vec{a}, \vec{b}, \vec{c}$ form a system of orthogonal vectors

$\frac{3}{2} < | \vec{a} + \vec{b} + \vec{c} | < \frac{5}{2}$

$| \vec{a} + \vec{b} | > | \vec{b} + \vec{c} |$

$| \vec{a} + 2 \vec{b} + 3 \vec{c} | < 20$

answer is A, C.

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Detailed Solution

${| \vec{a} - \vec{b} |}^{2} + {| \vec{b} - \vec{c} |}^{2} + {| \vec{c} + \vec{a} |}^{2} = 3 \Rightarrow {| \vec{a} + \vec{c} - \vec{b} |}^{2} = 0, \vec{a} + \vec{c} = \vec{b} \Rightarrow \vec{a} \cdot \vec{c} = - \frac{1}{2}$

So , $| \vec{a} + \vec{b} + \vec{c} | = 2 | \vec{b} | = 2$ , $| \vec{a} + 2 \vec{b} + 3 \vec{c} | = | 3 \vec{a} + 5 \vec{c} | = \sqrt{19}$

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Let a→ , b→ and c→ are three unit vectors such that |a→−b→|2+|b→−c→|2+|c→+a→|2=3 , then which of following option(s) is/are correct?

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Let $\vec{a}$ , $\vec{b}$ and $\vec{c}$ are three unit vectors such that ${| \vec{a} - \vec{b} |}^{2} + {| \vec{b} - \vec{c} |}^{2} + {| \vec{c} + \vec{a} |}^{2} = 3$ , then which of following option(s) is/are correct?