Miscellaneous · Q1: Unit vector in the XY-plane at 30° to the positive x-axis

Basics you need

Solution

1. A vector in the XY-plane has no $\hat k$ part

   Draw the vector from the origin O with its tip at P. Since it lies in the XY-plane, P has z-coordinate 0, so

   $$\vec r = \overrightarrow{OP} = x\hat i + y\hat j, \qquad |\vec r| = 1.$$
   The XY-plane is the plane $z = 0$⁠. Every point of it has the form $(x, y, 0)$⁠, so the $\hat k$ component of a vector lying in it is 0. We only have to find $x$ and $y$⁠.
   Position vector in component form (Section 10.5.1). Basics: three numbers fix a point

2. Read the components from the right triangle

   Let P′ be the foot of the perpendicular from P to the x-axis. In the right triangle OP′P the hypotenuse is $\text{OP} = |\vec r| = 1$ and $\angle \text{P}'\text{OP} = 30^\circ$⁠. So

   $$\begin{aligned} x &= \text{OP}' = 1 \cdot \cos 30^\circ, \\ y &= \text{P}'\text{P} = 1 \cdot \sin 30^\circ. \end{aligned}$$
   The $x$ component is the "shadow" of OP on the x-axis, the side next to the 30° angle, so it is hypotenuse $\times \cos 30^\circ$⁠. The $y$ component is the height of P above the axis, the side opposite the angle, so it is hypotenuse $\times \sin 30^\circ$⁠. The angle is measured from OX towards OY, the usual (anticlockwise) way, so P is above the x-axis and $y > 0$⁠.
   Sides of a right triangle from the hypotenuse and an angle. Basics: cos θ is a shadow, sin θ is a height

3. Put in the values

   With $\cos 30^\circ = \frac{\sqrt3}{2}$ and $\sin 30^\circ = \frac12$⁠,

   $$\vec r = \frac{\sqrt3}{2}\hat i + \frac12\hat j.$$
   These are exact values from the 30-60-90 triangle: half of an equilateral triangle of side 1 has sides $\frac12$⁠, $\frac{\sqrt3}{2}$ and 1. In general, a unit vector making angle $\theta$ with OX is $\cos\theta\,\hat i + \sin\theta\,\hat j$⁠.
   All unit vectors in the XY-plane: Example 26. Lesson: Examples 26–30

4. Check the length and the angle

   $$\begin{aligned} |\vec r| &= \sqrt{\left(\tfrac{\sqrt3}{2}\right)^2 + \left(\tfrac12\right)^2} \\ &= \sqrt{\tfrac34 + \tfrac14} = \sqrt1 = 1. \end{aligned}$$

   The cosine of the angle with OX is the $\hat i$ component of a unit vector: $\cos\alpha = \frac{\sqrt3}{2}$⁠, so $\alpha = 30^\circ$⁠.

   A unit vector's components are its direction cosines, so the $\hat i$ component $\frac{\sqrt3}{2}$ is $\cos$ of the angle with OX. The angle between two vectors lies in $[0^\circ, 180^\circ]$⁠, and $\cos\alpha = \frac{\sqrt3}{2}$ there gives only $\alpha = 30^\circ$⁠.
   Direction cosines are the components of the unit vector (Section 10.5.1, Remarks (iii)). Lesson: components

5. Is there another answer?

   The mirror image $\frac{\sqrt3}{2}\hat i - \frac12\hat j$ (dashed in the figure) is also a unit vector in the XY-plane, and its $\hat i$ component is also $\frac{\sqrt3}{2}$⁠, so it too makes an angle of 30° with OX. It lies below the x-axis: the 30° is turned clockwise, from OX away from OY.

   An angle between two vectors carries no sense of turning, so both vectors fit the words. The usual reading turns from OX towards OY, which gives $\frac{\sqrt3}{2}\hat i + \frac12\hat j$⁠; this is the vector the answer key gives. "Write down a unit vector" asks for one, so either is a correct example, and the anticlockwise one is the expected answer.
   Angle between two vectors, $0 \le \theta \le \pi$ (Section 10.6.1). Lesson: the dot product

Every fact used, and where it comes from