| Statement I: If \[\vec{a},\vec{b}\]and \[\vec{c}\] are the unit vectors such that \[\vec{a}+\vec{b}+\vec{c}=\vec{0}\]then \[\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}\] \[=-3/2\]. |
| Statement II: \[{{(\vec{x}+\vec{y})}^{2}}=|\vec{x}{{|}^{2}}+|\vec{y}{{|}^{2}}+2(\vec{x}.\vec{y})\] |
A) Statement I is true; Statement II is true; Statement II is not a correct explanation for Statement I,
B) Statement I is true; Statement II is false.
C) Statement I is false; Statement II is true.
D) Statement I is true; Statement II is true; Statements is the correct explanation for Statement I.
Correct Answer: D
Solution :
Given that, \[|\overrightarrow{a}|=1,|\overrightarrow{b}|=1,|\overrightarrow{c}|=1.\] \[\therefore \]\[|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{{|}^{2}}=|\overrightarrow{a}{{|}^{2}}+|\overrightarrow{b}{{|}^{2}}+|\overrightarrow{c}{{|}^{2}}\] \[+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})\] \[\Rightarrow \]\[0=1+1+1+(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})\] \[\therefore \]\[\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}=-3/2\] Hence, Statements I, II are true and Statement II is the correct explanation for Statement I.
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