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: Using the formula for the angle between two vectors.
The angle \( \theta \) between two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by the formula:
\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \] where: - \( \mathbf{A} = -2\hat{i} + \hat{j} + 3\hat{k} \) - \( \mathbf{B} = 3\hat{i} - 2\hat{j} + \hat{k} \) Step 2: Find the dot product \( \mathbf{A} \cdot \mathbf{B} \).
The dot product is calculated as:
\[ \mathbf{A} \cdot \mathbf{B} = (-2)(3) + (1)(-2) + (3)(1) \] \[ \mathbf{A} \cdot \mathbf{B} = -6 - 2 + 3 = -5 \] Step 3: Find the magnitudes of \( \mathbf{A} \) and \( \mathbf{B} \).
The magnitude of \( \mathbf{A} \) is:
\[ |\mathbf{A}| = \sqrt{(-2)^2 + 1^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \] The magnitude of \( \mathbf{B} \) is:
\[ |\mathbf{B}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \] Step 4: Substitute values into the formula.
Now substitute the values of \( \mathbf{A} \cdot \mathbf{B} \), \( |\mathbf{A}| \), and \( |\mathbf{B}| \) into the formula:
\[ \cos \theta = \frac{-5}{\sqrt{14} \times \sqrt{14}} = \frac{-5}{14} \] Step 5: Find the angle \( \theta \).
Now, find the angle \( \theta \) using the inverse cosine function:
\[ \theta = \cos^{-1} \left( \frac{-5}{14} \right) \] Using a calculator, we get:
\[ \theta \approx 112.62^\circ \] Final Answer:
The angle between the vectors is approximately \( 112.62^\circ \).