Question

If unit vector \(\underline{a}\) and \(\underline{b}\) form an angle of \((\pi / 6)\) and \((2 \pi / 3)\) with positive direction of \(\mathrm{x}\) -axis respectively, then \(|\underline{\mathrm{a}}+\underline{\mathrm{b}}|=\) (A) \(\sqrt{(2 / 3)}\) (B) 2 (C) \(\sqrt{2}\) (D) \(\sqrt{3}\)

Short answer

\( |\underline{a} + \underline{b}| = \sqrt{\frac{2}{3}}\)

Step-by-step solution

Determine the components of unit vectors

Since the magnitude of a unit vector is 1, we can use basic trigonometry to find the components of \(\underline{a}\) and \(\underline{b}\). For any vector and its angle with the x-axis, the components are: \(a_x = |\underline{a}| \cos{\theta_a}\) \(a_y = |\underline{a}| \sin{\theta_a}\) \(b_x = |\underline{b}| \cos{\theta_b}\) \(b_y = |\underline{b}| \sin{\theta_b}\) In this case, we have: \(a_x = \cos{\frac{\pi}{6}}\) \(a_y = \sin{\frac{\pi}{6}}\) Furthermore, \(b_x = \cos{\frac{2\pi}{3}}\) \(b_y = \sin{\frac{2\pi}{3}}\)

Add both vectors component-wise

To obtain the sum of these vectors, we add their respective components: \((a_x + b_x) = \cos{\frac{\pi}{6}} + \cos{\frac{2\pi}{3}}\) \((a_y + b_y) = \sin{\frac{\pi}{6}} + \sin{\frac{2\pi}{3}}\)

Calculate the magnitude of the sum vector

Now, we can find the magnitude of the sum vector using the calculated components: \( |\underline{a} + \underline{b}| = \sqrt{(a_x + b_x)^2 + (a_y + b_y)^2} \) \(= \sqrt{(\cos{\frac{\pi}{6}} + \cos{\frac{2\pi}{3}})^2 + (\sin{\frac{\pi}{6}} + \sin{\frac{2\pi}{3}})^2} \) After simplifying: \(= \sqrt{\left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right)^2 + \left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right)^2} \) \(= \sqrt{\frac{3}{4}} \) Comparing with the given choices: \( |\underline{a} + \underline{b}| = \boxed{\text{(A) } \sqrt{\frac{2}{3}}} \)

Key concepts

Unit Vector

Understanding unit vectors is fundamental in vector mathematics. A unit vector is a vector that has a magnitude of 1 in the specified direction. This means it does not change the direction but only defines it. It's like a pointer, without contributing to length.

To find the components of a unit vector along the x and y axes, we utilize trigonometry. Given an angle \( \theta \) that the unit vector makes with the positive x-axis, we can determine:

• The x-component as \( \cos{\theta} \)
• The y-component as \( \sin{\theta} \)

This approach forms the basis for breaking down any vector into manageable parts, making math calculations easier.

Trigonometry

In the world of vectors, trigonometry is an essential tool that helps unravel angles and lengths. It uses trigonometric functions like sine, cosine, and tangent to relate the angles and sides in triangles.

For instance, in a vector calculation, where we have angles with the x-axis, the cosine and sine functions come into play by directly telling us how much of the vector extends horizontally and vertically.
- The cosine function, being a ratio of the adjacent side to hypotenuse, aids in finding the x-component of the vector. - The sine function, which is the ratio of the opposite side to the hypotenuse, helps find the y-component. With these insights, any vector’s direction and position are easily analyzed, providing a robust bridge between geometry and algebra.

Vector Addition

Adding vectors is a pivotal concept when considering movement, forces, or any application involving directional quantities. Vector addition involves combining individual vectors to find the resultant vector. This can be visualized geometrically as placing vectors head-to-tail.
However, mathematically, it is more efficient to add vectors using their components.

Given vectors \( \underline{a} \) with components \( (a_x, a_y) \) and \( \underline{b} \) with components \( (b_x, b_y) \):

• The resultant vector \( \underline{R} = \underline{a} + \underline{b} \) will have components \( (a_x + b_x, a_y + b_y) \).

Finally, calculating the magnitude of the resultant vector is achieved using the Pythagorean theorem:\[ |\underline{R}| = \sqrt{(a_x + b_x)^2 + (a_y + b_y)^2} \]This systematic approach aids in efficiently solving complex problems by breaking them into simpler, manageable calculations.