Step 1: {Recall the dot product formula}
The dot product of two vectors \( \vec{a} \) and \( \vec{k} \) is: \[ |\vec{a} \cdot \vec{k}| = |\vec{a}| |\vec{k}| \cos \theta, \] where \( \theta \) is the angle between the vectors.
Step 2: {Determine the range of \( |\vec{a} \cdot \vec{k}| \)}
Since \( |\vec{a}| = 2 \), the magnitude of \( \vec{k} \) varies as: \[ |\vec{k}| \in [-3, 2]. \] The maximum value of \( |\vec{a} \cdot \vec{k}| \) occurs when \( \cos \theta = 1 \): \[ |\vec{a} \cdot \vec{k}|_{{max}} = 2 \cdot 3 = 6. \] The minimum value of \( |\vec{a} \cdot \vec{k}| \) occurs when \( \cos \theta = 0 \): \[ |\vec{a} \cdot \vec{k}|_{{min}} = 0. \]
Step 3: {Conclude the result}
Thus, \( |\vec{a} \cdot \vec{k}| \in [0, 6] \).
Show that the vectors \( 2\hat{i} - \hat{j} + \hat{k}, \hat{i} - 3\hat{j} - 5\hat{k}, 3\hat{i} - 4\hat{j} - 4\hat{k} \) form the vertices of a right-angled triangle.
Find the angle between the lines \[ \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z-5}{-5} \quad \text{and} \quad \frac{x+3}{-3} = \frac{y-1}{2} = \frac{z-5}{5}. \]
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: