Question

If $\overrightarrow a ,\overrightarrow b $ are the unit vectors inclined to x-axis at the angle $30^\circ {\text{ and 120}}^\circ $ then $\left| {\overrightarrow a + \overrightarrow b } \right|$ equals
A. $\sqrt {2/3} $
B. $\sqrt 2 $
C. $\sqrt 3 $
D. $2$

Answer 1

Hint:
We can write $\overrightarrow a ,\overrightarrow b $ in the form of $\widehat i,\widehat j$ and we know that $\overrightarrow a ,\overrightarrow b $ are unit vectors so we can say that
$\left| a \right| = 1,\left| b \right| = 1$
So let us assume that $\overrightarrow a = {x_1}\widehat i + {y_1}\widehat j$ and $\overrightarrow b = {x_2}\widehat i + {y_2}\widehat j$

Complete step by step solution:
Here we know that $\overrightarrow a ,\overrightarrow b $ are unit vectors so we can say that their magnitudes are equal to one
$\left| a \right| = 1,\left| b \right| = 1$
So let us assume that $\overrightarrow a = {x_1}\widehat i + {y_1}\widehat j$ and $\overrightarrow b = {x_2}\widehat i + {y_2}\widehat j$ are two vectors and we know that $\left| a \right| = 1,\left| b \right| = 1$ so we get that $\sqrt {{x_1}^2 + {y_1}^2} = 1$ and $\sqrt {{x_2}^2 + {y_2}^2} = 1$
Now we are also given that $\overrightarrow a ,\overrightarrow b $ are the unit vectors inclined to x-axis at the angle $30^\circ {\text{ and 120}}^\circ $
So we get the graph as

Therefore the angle made by the $\overrightarrow b $ by the negative x-axis is $180 - 120 = 60^\circ $
Now as we assumed that $\overrightarrow a = {x_1}\widehat i + {y_1}\widehat j$ so we can write that
${x_1} = \left| a \right|\cos 30^\circ ,{y_1} = \left| a \right|\sin 30^\circ $ and we know that $\left| a \right| = 1,\left| b \right| = 1$
So ${x_1} = \dfrac{{\sqrt 3 }}{2},{y_1} = \dfrac{1}{2}$
So we get the $\overrightarrow a = \dfrac{{\sqrt 3 }}{2}\widehat i + \dfrac{1}{2}\widehat j$
Now as we assumed that $\overrightarrow b = {x_2}\widehat i + {y_2}\widehat j$ so we can write that
${x_2} = - \left| b \right|\cos 60^\circ ,{y_2} = \left| b \right|\sin 60^\circ $ and we know that $\left| a \right| = 1,\left| b \right| = 1$
So ${x_2} = \dfrac{{ - 1}}{2},{y_2} = \dfrac{{\sqrt 3 }}{2}$
So we get the $\overrightarrow b = \dfrac{1}{2}\widehat i + \dfrac{{\sqrt 3 }}{2}\widehat j$
So we get
$\overrightarrow a + \overrightarrow b = \dfrac{{\sqrt 3 }}{2}\widehat i + \dfrac{1}{2}\widehat j$$ + \dfrac{1}{2}\widehat i + \dfrac{{\sqrt 3 }}{2}\widehat j$
$\overrightarrow a + \overrightarrow b = (\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{2})\widehat i + (\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2})\widehat j$
Now the magnitude of this can be written as
$\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt {{{\left( {\dfrac{{\sqrt 3 - 1}}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 + 1}}{2}} \right)}^2}} $
$\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt {\left( {\dfrac{{{{(\sqrt 3 - 1)}^2}}}{4}} \right) + \left( {\dfrac{{{{(\sqrt 3 + 1)}^2}}}{4}} \right)} $
$\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt {\dfrac{{3 + 1 + 3 + 1 - 2\sqrt 3 + 2\sqrt 3 }}{4}} = \sqrt {\dfrac{8}{4}} = \sqrt 2 $

So we got that $\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt 2 $

Note:
As we know that $\left| a \right| = 1,\left| b \right| = 1$ and the angle between them is $\theta = 120 - 30 = 90^\circ $
So $\overrightarrow a + \overrightarrow b $ is the resultant of $\overrightarrow a {\text{ and }}\overrightarrow b $ and its magnitude is given as
$\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt {{{\left( {\left| {\overrightarrow a } \right|} \right)}^2} + {{\left( {\left| {\overrightarrow b } \right|} \right)}^2} + 2\left( {\left| {\overrightarrow a } \right|} \right)\left( {\left| {\overrightarrow b } \right|} \right)\cos \theta } $
$
   = \sqrt {{1^2} + {1^2} + 2.1.1.\cos 90} \\
   = \sqrt 2 \\
 $