Problem on vector length
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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
Definition of a unit vector
A unit vector is a vector of length 1.
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
Definition of vector length
The length of $x_1 \mathbf{i} + x_2 \mathbf{j} + x_3 \mathbf{k}$ is $\sqrt{x_1^2 + x_2^2 + x_3^2}$.
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
Length of a scalar vector multiplication
For any scalar $c$ and vector $\mathbf{x}$, the length $|c \mathbf{x} | = |c|\ |\mathbf{x} |$.
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
Definition of vector length
The length of $x_1 \mathbf{i} + x_2 \mathbf{j} + x_3 \mathbf{k}$ is $\sqrt{x_1^2 + x_2^2 + x_3^2}$.
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 $c = \pm \frac{1}{\sqrt{3}}$.
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