Understanding Unit Vectors and Angles

The problem asks us to find the magnitude of the sum of two unit vectors, \(\overrightarrow a\) and \(\overrightarrow b\), given their angles of inclination with the x-axis. A unit vector is a vector with a magnitude of 1. Understanding how to represent unit vectors using angles is key.

Given:

• \(\overrightarrow a\) is a unit vector inclined at 30° to the x-axis.
• \(\overrightarrow b\) is a unit vector inclined at 120° to the x-axis.
• We need to find \(\left| {\overrightarrow a + \overrightarrow b } \right|\), the magnitude of the sum of these unit vectors.

Representing Unit Vectors in Component Form

A vector \(\overrightarrow v\) with magnitude \(|\overrightarrow v|\) inclined at an angle \(\theta\) to the x-axis can be written in component form as \(\overrightarrow v = |\overrightarrow v| \cos \theta \hat{i} + |\overrightarrow v| \sin \theta \hat{j}\). Since both \(\overrightarrow a\) and \(\overrightarrow b\) are unit vectors, their magnitudes are 1.

• For \(\overrightarrow a\): Angle \(\theta_a = 30^\circ\), \(|\overrightarrow a| = 1\).
  \(\overrightarrow a = 1 \cdot \cos 30^\circ \hat{i} + 1 \cdot \sin 30^\circ \hat{j}\)
  \(\overrightarrow a = \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}\)
• For \(\overrightarrow b\): Angle \(\theta_b = 120^\circ\), \(|\overrightarrow b| = 1\).
  \(\overrightarrow b = 1 \cdot \cos 120^\circ \hat{i} + 1 \cdot \sin 120^\circ \hat{j}\)
  We know \(\cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2}\) and \(\sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}\).
  \(\overrightarrow b = -\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}\)

Calculating the Sum of the Unit Vectors

Now we add the two unit vectors component-wise:

\(\overrightarrow a + \overrightarrow b = \left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) + \left( -\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j} \right)

\(\overrightarrow a + \overrightarrow b = \left( \frac{\sqrt{3}}{2} - \frac{1}{2} \right) \hat{i} + \left( \frac{1}{2} + \frac{\sqrt{3}}{2} \right) \hat{j}

\(\overrightarrow a + \overrightarrow b = \frac{\sqrt{3}-1}{2} \hat{i} + \frac{1+\sqrt{3}}{2} \hat{j}\)

Finding the Vector Magnitude

The magnitude of a vector \(c \hat{i} + d \hat{j}\) is given by \(\sqrt{c^2 + d^2}\). For the sum vector \(\overrightarrow a + \overrightarrow b\), the components are \(c = \frac{\sqrt{3}-1}{2}\) and \(d = \frac{1+\sqrt{3}}{2}\). We need to find the vector magnitude \(\left| {\overrightarrow a + \overrightarrow b } \right|\).

\(\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt{ \left( \frac{\sqrt{3}-1}{2} \right)^2 + \left( \frac{1+\sqrt{3}}{2} \right)^2 }

\(\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt{ \frac{(\sqrt{3})^2 - 2\sqrt{3}(1) + 1^2}{4} + \frac{1^2 + 2(1)\sqrt{3} + (\sqrt{3})^2}{4} }

\(\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt{ \frac{3 - 2\sqrt{3} + 1}{4} + \frac{1 + 2\sqrt{3} + 3}{4} }

\(\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt{ \frac{4 - 2\sqrt{3}}{4} + \frac{4 + 2\sqrt{3}}{4} }

\(\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt{ \frac{(4 - 2\sqrt{3}) + (4 + 2\sqrt{3})}{4} }

\(\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt{ \frac{8}{4} }\)

\(\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt{2}\)

Alternative Method: Using the Angle Between Unit Vectors

We can also find the magnitude of the sum of two vectors using the formula \(\left| {\overrightarrow u + \overrightarrow v } \right| = \sqrt{|\overrightarrow u|^2 + |\overrightarrow v|^2 + 2|\overrightarrow u||\overrightarrow v|\cos \theta}\), where \(\theta\) is the angle between the vectors. For our unit vectors \(\overrightarrow a\) and \(\overrightarrow b\), the magnitudes are \(|\overrightarrow a|=1\) and \(|\overrightarrow b|=1\).

The angle between \(\overrightarrow a\) and \(\overrightarrow b\) is the difference between their angles with the x-axis: \(\theta = 120^\circ - 30^\circ = 90^\circ\).

Using the formula for the magnitude of the sum of vectors:

\(\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt{|a|^2 + |b|^2 + 2|a||b|\cos \theta}\)

\(\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt{1^2 + 1^2 + 2(1)(1)\cos 90^\circ}\)

Since \(\cos 90^\circ = 0\):

\(\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt{1 + 1 + 2(1)(1)(0)}\)

\(\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt{2 + 0}\)

\(\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt{2}\)

Both methods confirm that the magnitude of the sum of the unit vectors is \(\sqrt{2}\).