Example 6: Unit vector along $2\hat i+3\hat j+\hat k$

Basics you need

Solution

1. Find the magnitude

   $$\begin{aligned} |\vec a| &= \sqrt{2^2 + 3^2 + 1^2} \\ &= \sqrt{4 + 9 + 1} = \sqrt{14}. \end{aligned}$$
   The magnitude of $x\hat i + y\hat j + z\hat k$ is the distance from O to the point $(x, y, z)$⁠. Here the point is A(2, 3, 1). Since $|\vec a| = \sqrt{14} \ne 0$⁠, $\vec a$ is not the zero vector, so it has a direction and a unit vector along it exists.
   $|\vec r| = \sqrt{x^2 + y^2 + z^2}$ (Section 10.5.1). Basics: use Pythagoras twice

2. Divide by the magnitude

   $$\hat a = \frac{1}{|\vec a|}\vec a = \frac{1}{\sqrt{14}}(2\hat i + 3\hat j + \hat k).$$
   With $\lambda = \frac{1}{|\vec a|} > 0$⁠, the vector $\lambda\vec a$ has the same direction as $\vec a$ and magnitude $\lambda|\vec a| = \frac{1}{|\vec a|}\cdot|\vec a| = 1$⁠. That is exactly what "unit vector in the direction of $\vec a$" asks for.
   $\hat a = \frac{1}{|\vec a|}\vec a$ (Section 10.5). Lesson: components

3. Distribute over the components

   $$\hat a = \frac{2}{\sqrt{14}}\hat i + \frac{3}{\sqrt{14}}\hat j + \frac{1}{\sqrt{14}}\hat k.$$

   This is the required unit vector.

   Multiplying a vector by a scalar multiplies each component by that scalar. The answer may also be written $\frac{\sqrt{14}}{14}(2\hat i + 3\hat j + \hat k)$ after rationalising; both forms are the same vector.
   $\lambda\vec a = (\lambda a_1)\hat i + (\lambda a_2)\hat j + (\lambda a_3)\hat k$ (Section 10.5.1 (iv)). Lesson: components

4. Check the length and the direction

   $$|\hat a|^2 = \frac{4}{14} + \frac{9}{14} + \frac{1}{14} = \frac{14}{14} = 1.$$

   So $|\hat a| = 1$⁠, and $\hat a = \frac{1}{\sqrt{14}}\vec a$ is a positive multiple of $\vec a$⁠, so it points the same way.

   These two checks are the two halves of the definition: length 1, and the direction of $\vec a$⁠. A common slip (dividing by 14 instead of $\sqrt{14}$⁠) is caught at once, because then the squares add to $\frac{14}{196} = \frac{1}{14}$⁠, not 1.
   A unit vector has magnitude 1 (Section 10.3); $|\lambda\vec a| = |\lambda||\vec a|$ (Section 10.5). Lesson: six kinds of vectors worth naming

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