A) \[4\sqrt{5}\]
B) \[4\sqrt{3}\]
C) \[4\sqrt{17}\]
D) None of the above
Correct Answer: B
Solution :
Diagonal \[\overrightarrow{\mathbf{AC}}=\overset{\to }{\mathop{\mathbf{a}}}\,+\overset{\to }{\mathop{\mathbf{b}}}\,\] \[\Rightarrow \] \[\left| \overrightarrow{\mathbf{AC}} \right|=\left| \overset{\to }{\mathop{\mathbf{a}}}\,+\overset{\to }{\mathop{\mathbf{b}}}\, \right|\] \[\Rightarrow \] \[\left| \overrightarrow{\mathbf{AC}} \right|=\left| \overset{\to }{\mathop{\mathbf{a}}}\, \right|+\left| \overset{\to }{\mathop{\mathbf{b}}}\, \right|+2\overset{\to }{\mathop{\mathbf{a}}}\,\cdot \overset{\to }{\mathop{\mathbf{b}}}\,\] \[\Rightarrow \] \[{{\left| \overrightarrow{\mathbf{AC}} \right|}^{2}}=\{|3\overset{\to }{\mathop{\alpha }}\,-\overset{\to }{\mathop{\beta }}\,|+|\overset{\to }{\mathop{\alpha }}\,+3\overset{\to }{\mathop{\beta }}\,|\] \[+2(3\overset{\to }{\mathop{\alpha }}\,-\overset{\to }{\mathop{\beta }}\,)\cdot (\overset{\to }{\mathop{\alpha }}\,+3\overset{\to }{\mathop{\beta }}\,)\}\] \[=9\overset{\to }{\mathop{\alpha \,_{{}}^{2}}}\,+\overset{\to }{\mathop{\beta \,_{{}}^{2}}}\,-6\alpha \cdot \beta +\overset{\to }{\mathop{\alpha \,_{{}}^{2}}}\,+9\overset{\to }{\mathop{\beta \,_{{}}^{2}}}\,+6\overset{\to }{\mathop{\alpha }}\,\cdot \overset{\to }{\mathop{\beta }}\,\] \[+6\overset{\to }{\mathop{\alpha \,_{{}}^{2}}}\,-6\overset{\to }{\mathop{\beta \,_{{}}^{2}}}\,+16\overset{\to }{\mathop{\alpha }}\,\cdot \overset{\to }{\mathop{\beta }}\,\] \[\Rightarrow \] \[{{\left| \overrightarrow{\mathbf{AC}} \right|}^{2}}=16\overset{\to }{\mathop{\alpha \,_{{}}^{2}}}\,+4\overset{\to }{\mathop{\beta \,_{{}}^{2}}}\,+16\overset{\to }{\mathop{\alpha }}\,\cdot \overset{\to }{\mathop{\beta }}\,\] \[\Rightarrow \] \[{{\left| \overrightarrow{\mathbf{AC}} \right|}^{2}}=64+16|\overset{\to }{\mathop{\alpha }}\,||\overset{\to }{\mathop{\beta }}\,|cos\frac{\pi }{3}\] \[{{\left| \overrightarrow{\mathbf{AC}} \right|}^{2}}=80+16\times 4\times \frac{1}{2}=112\] \[\Rightarrow \] \[{{\left| \overrightarrow{\mathbf{AC}} \right|}^{2}}=\sqrt{112}=4\sqrt{7}\] Other diagonal is \[{{\left| \overrightarrow{\mathbf{BD}} \right|}^{2}}=|\overset{\to }{\mathop{\mathbf{a}}}\,-\overset{\to }{\mathop{\mathbf{b}}}\,|\] \[{{\left| \overrightarrow{\mathbf{BD}} \right|}^{2}}=|\overset{\to }{\mathop{\mathbf{a}}}\,{{|}^{2}}|\overset{\to }{\mathop{\mathbf{b}}}\,{{|}^{2}}-2|\overset{\to }{\mathop{\mathbf{a}}}\,|\cdot |\overset{\to }{\mathop{\mathbf{b}}}\,|\] \[=64+16-16\times 4\times \frac{1}{2}\] \[=80-32\] \[=48\] \[\Rightarrow \] \[|\overrightarrow{\mathbf{BD}}|=\sqrt{48}=4\sqrt{3}\]You need to login to perform this action.
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