14. Find the scalar \( t \) (or show that there is none) so that the vector \( \boldsymbol{v}=t \boldsymbol{i}-2 t \boldsymbol{j}+3 t \boldsymbol{k} \) is a unit vector.

A unit vector is defined as a vector with a magnitude of 1. To find the scalar \( t \) for the vector \( \boldsymbol{v} = t \boldsymbol{i} - 2t \boldsymbol{j} + 3t \boldsymbol{k} \) to be a unit vector, we need to set the magnitude of \( \boldsymbol{v} \) equal to 1. The magnitude of \( \boldsymbol{v} \) can be calculated using the formula: \[ |\boldsymbol{v}| = \sqrt{t^2 + (-2t)^2 + (3t)^2} \] Calculating this gives us: \[ |\boldsymbol{v}| = \sqrt{t^2 + 4t^2 + 9t^2} = \sqrt{14t^2} = \sqrt{14} |t| \] Setting this equal to 1, we have: \[ \sqrt{14} |t| = 1 \] Now we can solve for \( |t| \): \[ |t| = \frac{1}{\sqrt{14}} \quad \Rightarrow \quad t = \pm \frac{1}{\sqrt{14}} \] Thus, there are two possible values for \( t \) that will make \( \boldsymbol{v} \) a unit vector: \[ t = \frac{1}{\sqrt{14}} \quad \text{or} \quad t = -\frac{1}{\sqrt{14}} \]