Question
The angles of a triangle, two of whose sides are represented by the vectors $$\sqrt 3 \left( {\overrightarrow a \times \overrightarrow b } \right)$$ and $$\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow a $$ where $$\overrightarrow b $$ is a non-zero vector and $$\overrightarrow a $$ is a unit vector are :
A.
$${\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right);\,{\tan ^{ - 1}}\left( {\frac{1}{2}} \right);\,{\tan ^{ - 1}}\left( {\frac{{\sqrt 3 + 2}}{{1 - 2\sqrt 3 }}} \right)$$
B.
$${\tan ^{ - 1}}\left( {\sqrt 3 } \right);\,{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right);\,{\cot ^{ - 1}}\left( 0 \right)$$
C.
$${\tan ^{ - 1}}\left( {\sqrt 3 } \right);\,{\tan ^{ - 1}}\left( 2 \right);\,{\tan ^{ - 1}}\left( {\frac{{\sqrt 3 + 2}}{{2\sqrt 3 - 1}}} \right)$$
D.
$${\tan ^{ - 1}}\left( {\sqrt 3 } \right);\,{\tan ^{ - 1}}\left( {\sqrt 2 } \right);\,{\tan ^{ - 1}}\left( {\frac{{\sqrt 2 + 3}}{{3\sqrt 2 - 1}}} \right)$$
Answer :
$${\tan ^{ - 1}}\left( {\sqrt 3 } \right);\,{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right);\,{\cot ^{ - 1}}\left( 0 \right)$$
Solution :
Let $$ABC$$ be a triangle in which the given vectors are represented by the sides $$AB$$ and $$AC$$
$$\eqalign{
& {\text{i}}{\text{.e}}{\text{., }}AB = \sqrt 3 \left( {a \times b} \right) \cr
& {\text{and }}AC = b - \left( {a.b} \right)a \cr
& \therefore \,AB.AC = \sqrt {3\left( {a \times b} \right)\left[ {b - \left( {a.b} \right)a} \right]} \cr
& = \sqrt {3\left[ {\left( {a \times b} \right).b\left( {a.b} \right)\left( {a \times b} \right)a} \right]} \cr
& = \sqrt {3\left( {0 - 0} \right)} \cr
& = 0 \cr
& {\text{Therefore, }}\angle BAC = {90^ \circ } \cr
& A{B^2} = {\left| {\sqrt 3 \left( {a \times b} \right)} \right|^2} = 3{\left( {a \times b} \right)^2} \cr
& A{C^2} = {\left[ {b - \left( {a.b} \right)a} \right]^2}......\left( {\text{i}} \right) \cr
& = {\left( b \right)^2} + {\left( {a.b} \right)^2}{a^2} - 2\left( {b.a} \right)\left( {a.b} \right) \cr
& = {\left( b \right)^2} + {\left( {a.b} \right)^2} - 2{\left( {a.b} \right)^2} \cr
& = {\left( b \right)^2} - {\left( {a.b} \right)^2} = {\left( b \right)^2} = {\left| a \right|^2}{\left| b \right|^2}{\cos ^2}\theta \cr
& = {\left( b \right)^2}\left[ {1 - {{\left| a \right|}^2}{{\cos }^2}\theta } \right] \cr
& = {\left( b \right)^2}\left( {1 - {{\cos }^2}\theta } \right) \cr
& = {\left( b \right)^2}{\sin ^2}\theta \cr
& = {\left| a \right|^2}{\left| b \right|^2}{\sin ^2}\theta \cr
& = {\left( {a \times b} \right)^2}......\left( {{\text{ii}}} \right) \cr
& {\text{Dividing equation }}\left( {\text{i}} \right){\text{ by }}\left( {{\text{ii}}} \right){\text{ we get}} \cr
& \frac{{A{B^2}}}{{A{C^2}}} = \frac{{3{{\left( {a \times b} \right)}^2}}}{{{{\left( {a \times b} \right)}^2}}} \cr
& \Rightarrow A{B^2} = 3.A{C^2} \cr
& \Rightarrow AB = \sqrt 3 AC \cr
& \tan \,C = \frac{{AB}}{{AC}} = \frac{{\sqrt 3 AC}}{{AC}} = \sqrt 3 \cr
& \therefore \,\angle C = {60^ \circ } \cr
& \therefore \,\angle A = {180^ \circ } - {90^ \circ } - {60^ \circ } = {30^ \circ } \cr
& {\text{Hence, angles of the triangle are 3}}{0^ \circ },\,{90^ \circ }{\text{ and }}{60^ \circ } \cr} $$