A) \[|\vec{a}|=2|\vec{b}|\]
B) \[2|\vec{a}|=|\vec{b}|\]
C) \[|\vec{a}|=\sqrt{3}|\vec{b}|\]
D) \[|\vec{b}|=\sqrt{3}|\vec{a}|\]
Correct Answer: D
Solution :
[d] \[(\overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,).\overset{\to }{\mathop{a}}\,=|\overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,||\overset{\to }{\mathop{a}}\,|cos\,\,60{}^\circ \] |
\[\cos \,60{}^\circ =\frac{1}{2}=\frac{(\overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,).\overset{\to }{\mathop{a}}\,}{\left| \overrightarrow{a}+\overrightarrow{b} \right|\left| \overrightarrow{a} \right|}=\frac{{{\left| a \right|}^{2}}}{\left| \overrightarrow{a}+\overrightarrow{b} \right|\left| \overrightarrow{a} \right|}\] |
\[\frac{1}{2}=\frac{\left| \overset{\to }{\mathop{a}}\, \right|}{\left| \overrightarrow{a}+\overrightarrow{b} \right|}\] ?(1) |
\[\left\{ as\,\,\overrightarrow{a}.\overrightarrow{b}=0,\overrightarrow{a}\bot \overrightarrow{b} \right\}\] |
\[\left( \overrightarrow{a}+\overrightarrow{b} \right).\overrightarrow{b}=\left| \overrightarrow{a}+\overrightarrow{b} \right|\left| \overrightarrow{b} \right|\cos \,\,30{}^\circ \] |
\[\cos \,\,30{}^\circ =\frac{\sqrt{3}}{2}=\frac{\left( \overrightarrow{a}+\overrightarrow{b} \right).\left( \overrightarrow{b} \right)}{\left| \overrightarrow{a}+\overrightarrow{b} \right|\left| \overrightarrow{b} \right|}\] |
\[\frac{\sqrt{3}}{2}=\frac{{{\left| {\vec{b}} \right|}^{2}}}{\left| \overrightarrow{a}+\overrightarrow{b} \right|\left| \overrightarrow{b} \right|}=\frac{\left| {\vec{b}} \right|}{\left| \overrightarrow{a}+\overrightarrow{b} \right|}\] ?(2) |
\[\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\left| {\vec{b}} \right|}{\left| {\vec{a}} \right|}\sqrt{3}\frac{\left| {\vec{b}} \right|}{\left| {\vec{a}} \right|}\Rightarrow \left| {\vec{b}} \right|=\sqrt{3}\left| {\vec{a}} \right|\] |
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