Question

Find the angle between the vectors \(-2\hat{i} + \hat{j} + 3\hat{k}\) and \(3\hat{i} - 2\hat{j} + \hat{k}\).

Hint:

{Key Points:}

• Dot product gives cosine of angle
• Negative dot product means angle>90°
• \(|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}\)
• \(\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}\)

Answer 1

Step 1: Recall the formula for the angle between two vectors If \(\vec{a}\) and \(\vec{b}\) are two vectors, then the angle \(\theta\) between them is given by: \[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \, |\vec{b}|} \]
Step 2: Identify the vectors \[ \vec{a} = -2\hat{i} + \hat{j} + 3\hat{k} \] \[ \vec{b} = 3\hat{i} - 2\hat{j} + \hat{k} \]
Step 3: Calculate the dot product \(\vec{a} \cdot \vec{b}\) \[ \vec{a} \cdot \vec{b} = (-2)(3) + (1)(-2) + (3)(1) \] \[ \vec{a} \cdot \vec{b} = -6 - 2 + 3 \] \[ \vec{a} \cdot \vec{b} = -5 \]
Step 4: Calculate the magnitudes of \(\vec{a}\) and \(\vec{b}\) \[ |\vec{a}| = \sqrt{(-2)^2 + (1)^2 + (3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \] \[ |\vec{b}| = \sqrt{(3)^2 + (-2)^2 + (1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \]
Step 5: Apply the formula \[ \cos \theta = \frac{-5}{\sqrt{14} \times \sqrt{14}} = \frac{-5}{14} \]
Step 6: Find the angle \[ \theta = \cos^{-1} \left( \frac{-5}{14} \right) \] \[ \theta = \cos^{-1} \left( -0.3571 \right) \approx 111^\circ \ (\text{approximately}) \]
Step 7: Final answer \[ \boxed{\theta = \cos^{-1}\left(-\frac{5}{14}\right)} \] or approximately \(\boxed{111^\circ}\).