a and b are two non collinear unit vectors. Then a , b , x a -y b form a triangle, if:

Mathematics · JEE Advanced · NTA Exams — Three Dimensional Geometry

1. x = –1; y = 1 and \(|\vec{a}+\vec{b}|=2 \cos \left(\frac{\vec{a} \wedge \vec{b}}{2}\right)\)
2. x = –1; y = 1 and \(\cos (\overrightarrow{\mathrm{a}} \wedge \overrightarrow{\mathrm{~b}})+\) 
   \(|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}| \cos [\overrightarrow{\mathrm{a}} \wedge-(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})]=-1\)
3. \(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}} \left\lvert\,=-2 \cot \left(\frac{\overrightarrow{\mathrm{a}}^{\wedge} \overrightarrow{\mathrm{b}}}{2}\right) \cos \left(\frac{\overrightarrow{\mathrm{a}}^{\wedge} \overrightarrow{\mathrm{b}}}{2}\right)\right.\) and
   x = –1, y = 1
4. none of these

Answer

(A) x = –1; y = 1 and | a + b |=2 ( a b 2 ), (B) x = –1; y = 1 and ( a ~b )+ | a + b | [ a -( a + b )]=-1

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