Answer
The statement makes sense.
Work Step by Step
Assume the two vectors A and B subtend at angle $\theta $ such that;
$\mathbf{A}=\mathbf{i}-\mathbf{j}+\mathbf{k}$
And,
$\mathbf{B}=-\mathbf{i}+\mathbf{j}-\mathbf{k}$
Now consider the dot product of two vectors A and B.
$\begin{align}
& \mathbf{A}\cdot \mathbf{B}=\left( \mathbf{i}-\mathbf{j}+\mathbf{k} \right)\cdot \left( -\mathbf{i}+\mathbf{j}-\mathbf{k} \right) \\
& AB\cos \theta =-1-1-1 \\
& \sqrt{{{\left( 1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( 1 \right)}^{2}}}\times \sqrt{{{\left( 1 \right)}^{2}}+{{\left( 1 \right)}^{2}}+{{\left( -1 \right)}^{2}}}\cos \theta =-3 \\
& \sqrt{3}\times \sqrt{3}\cos \theta =-3
\end{align}$
Simplify further,
$\begin{align}
& 3\cos \theta =-3 \\
& \cos \theta =-\frac{3}{3} \\
& \cos \theta =-1 \\
& \theta =\pi
\end{align}$
Since, the angle between vectors A and B is $\theta =\pi $, hence when the dot product of two vectors is negative, the angle between them is obtuse.
Therefore the given statement makes sense.
