## Problem on vector length

For what value(s) of $c$ is $c (\textbf{i} + \textbf{j} + \textbf{k})$ a unit vector?
• ## Solution by direct computation

We begin by recalling that
Our strategy is to compute the length of $c ({\textbf{i}} + {\textbf{j}} + {\textbf{k}})$ and then find the value(s) of $c$ for which it is 1.
Recall that
The length of $c ({\textbf{i}} + {\textbf{j}} + {\textbf{k}})$ is $\sqrt{c^2 + c^2 + c^2} = \sqrt{3c^2} = \sqrt{3} |c|$.
Setting this length to be $1$, we get $c = \pm \frac{1}{\sqrt{3}}$.
• ## Solution using the length of a scalar-vector product

Recall that
The length of $c ({\textbf{i}} + {\textbf{j}} + {\textbf{k}})$ is $\vert c \vert \cdot \vert {\textbf{i}} + {\textbf{j}} + {\textbf{k}} \vert$.
In order to compute $\vert c \vert \cdot \vert {\textbf{i}} + {\textbf{j}} + {\textbf{k}} \vert$, we recall
We get, $|{\textbf{i}} + {\textbf{j}} + {\textbf{k}} |= \sqrt{3}$.
The length of $c ({\textbf{i}} + {\textbf{j}} + {\textbf{k}})$ is thus $|c| \sqrt{3}$.
This length is 1 when $c = \pm \frac{1}{\sqrt{3}}$.