JEE MAIN - Mathematics (2020 - 6th September Morning Slot - No. 8)
If $$\overrightarrow a $$
and $$\overrightarrow b $$
are unit vectors, then the greatest value of
$$\sqrt 3 \left| {\overrightarrow a + \overrightarrow b } \right| + \left| {\overrightarrow a - \overrightarrow b } \right|$$ is_____.
$$\sqrt 3 \left| {\overrightarrow a + \overrightarrow b } \right| + \left| {\overrightarrow a - \overrightarrow b } \right|$$ is_____.
Answer
4
Explanation
Let angle between $$\overrightarrow a $$ and $$\overrightarrow b $$
be $$\theta $$.
$$\sqrt 3 \left| {\overrightarrow a + \overrightarrow b } \right| + \left| {\overrightarrow a - \overrightarrow b } \right|$$
= $$\sqrt 3 \left( {\sqrt {1 + 1 + 2\cos \theta } } \right)$$ + $$\left( {\sqrt {1 + 1 - 2\cos \theta } } \right)$$
= $$\sqrt 3 \left( {\sqrt {2 + 2\cos \theta } } \right)$$ + $$\left( {\sqrt {2 - 2\cos \theta } } \right)$$
= $$\sqrt 6 \left( {\sqrt {1 + \cos \theta } } \right)$$ + $$\sqrt 2 \left( {\sqrt {1 - \cos \theta } } \right)$$
= $$\sqrt 6 \left( {\sqrt {2{{\cos }^2}{\theta \over 2}} } \right)$$ + $$\sqrt 2 \left( {\sqrt {2{{\sin }^2}{\theta \over 2}} } \right)$$
= $$2\sqrt 3 \left| {\cos {\theta \over 2}} \right|$$ + 2$$\left| {\sin {\theta \over 2}} \right|$$
$$ \le $$ $$\sqrt {{{\left( {2\sqrt 3 } \right)}^2} + {{\left( 2 \right)}^2}} $$ = 4
Note : |x| = $$\sqrt {{x^2}} $$
|x - 1| = $$\sqrt {{{\left( {x - 1} \right)}^2}} $$
|sin x| = $$\sqrt {{{\sin }^2}x} $$
That is why $${\sqrt {{{\sin }^2}{\theta \over 2}} }$$ = $$\left| {\sin {\theta \over 2}} \right|$$ and $${\sqrt {{{\cos }^2}{\theta \over 2}} }$$ = $$\left| {\cos {\theta \over 2}} \right|$$
$$\sqrt 3 \left| {\overrightarrow a + \overrightarrow b } \right| + \left| {\overrightarrow a - \overrightarrow b } \right|$$
= $$\sqrt 3 \left( {\sqrt {1 + 1 + 2\cos \theta } } \right)$$ + $$\left( {\sqrt {1 + 1 - 2\cos \theta } } \right)$$
= $$\sqrt 3 \left( {\sqrt {2 + 2\cos \theta } } \right)$$ + $$\left( {\sqrt {2 - 2\cos \theta } } \right)$$
= $$\sqrt 6 \left( {\sqrt {1 + \cos \theta } } \right)$$ + $$\sqrt 2 \left( {\sqrt {1 - \cos \theta } } \right)$$
= $$\sqrt 6 \left( {\sqrt {2{{\cos }^2}{\theta \over 2}} } \right)$$ + $$\sqrt 2 \left( {\sqrt {2{{\sin }^2}{\theta \over 2}} } \right)$$
= $$2\sqrt 3 \left| {\cos {\theta \over 2}} \right|$$ + 2$$\left| {\sin {\theta \over 2}} \right|$$
$$ \le $$ $$\sqrt {{{\left( {2\sqrt 3 } \right)}^2} + {{\left( 2 \right)}^2}} $$ = 4
Note : |x| = $$\sqrt {{x^2}} $$
|x - 1| = $$\sqrt {{{\left( {x - 1} \right)}^2}} $$
|sin x| = $$\sqrt {{{\sin }^2}x} $$
That is why $${\sqrt {{{\sin }^2}{\theta \over 2}} }$$ = $$\left| {\sin {\theta \over 2}} \right|$$ and $${\sqrt {{{\cos }^2}{\theta \over 2}} }$$ = $$\left| {\cos {\theta \over 2}} \right|$$
Comments (0)
