A) \[47\]
B) \[25\]
C) \[50\]
D) \[-25\]
Correct Answer: D
Solution :
Given that,\[|\overrightarrow{\mathbf{a}}|=3,\,\,|\overrightarrow{\mathbf{b}}|=4\]and\[|\overrightarrow{\mathbf{c}}|=5\] and \[\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=0\] On squaring both sides, we get \[|\overrightarrow{\mathbf{a}}{{|}^{2}}+|\overrightarrow{\mathbf{b}}{{|}^{2}}+|\overrightarrow{\mathbf{c}}{{|}^{2}}+2(\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}}\cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}}\cdot \overrightarrow{\mathbf{a}})=0\] \[\Rightarrow \] \[{{3}^{2}}+{{4}^{2}}+{{5}^{2}}+2(\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}}\cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}}\cdot \overrightarrow{\mathbf{a}})=0\] \[\Rightarrow \] \[2(\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}}\cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}}\cdot \overrightarrow{\mathbf{a}})=-(9+16+25)\] \[\Rightarrow \] \[\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}}\cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}}\cdot \overrightarrow{\mathbf{a}}=-\frac{50}{2}\] \[\Rightarrow \] \[\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}}\cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}}\cdot \overrightarrow{\mathbf{a}}=-25\]
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