question_answer

If \[\vec{a}.\,\vec{b}=0\] and \[\vec{a}+\vec{b}\] makes an angle of \[60{}^\circ \] with \[\vec{a}\], then

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)

\[\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\left| {\vec{b}} \right|}{\left| {\vec{a}} \right|}\sqrt{3}\frac{\left| {\vec{b}} \right|}{\left| {\vec{a}} \right|}\Rightarrow \left| {\vec{b}} \right|=\sqrt{3}\left| {\vec{a}} \right|\]