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:

When the dot product \( \vec{a} \cdot \vec{b} \) is negative, it implies that the angle \( \theta \) between the vectors is obtuse (\( 90^\circ<\theta \leq 180^\circ \)). If the dot product is positive, the angle is acute. If it is zero, the vectors are orthogonal (perpendicular). This can be a quick check on your answer.

Answer 1

Step 1: Understanding the Concept:
The angle \( \theta \) between two non-zero vectors \( \vec{a} \) and \( \vec{b} \) can be found using the scalar (dot) product formula. The dot product relates the angle between the vectors to the product of their magnitudes.
Step 2: Key Formula or Approach:
The dot product of two vectors \( \vec{a} \) and \( \vec{b} \) is defined as: \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta \] From this, we can find the angle \( \theta \) using: \[ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \] For vectors \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \):
\( \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \)
\( |\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)
\( |\vec{b}| = \sqrt{b_1^2 + b_2^2 + b_3^2} \)
Step 3: Detailed Calculation:
Let the given vectors be: \[ \vec{a} = -2\hat{i} + \hat{j} + 3\hat{k} \] \[ \vec{b} = 3\hat{i} - 2\hat{j} + \hat{k} \] 1. Calculate the dot product \( \vec{a} \cdot \vec{b} \): \[ \vec{a} \cdot \vec{b} = (-2)(3) + (1)(-2) + (3)(1) \] \[ \vec{a} \cdot \vec{b} = -6 - 2 + 3 = -5 \] 2. Calculate the magnitude of \( \vec{a} \): \[ |\vec{a}| = \sqrt{(-2)^2 + (1)^2 + (3)^2} \] \[ |\vec{a}| = \sqrt{4 + 1 + 9} = \sqrt{14} \] 3. Calculate the magnitude of \( \vec{b} \): \[ |\vec{b}| = \sqrt{(3)^2 + (-2)^2 + (1)^2} \] \[ |\vec{b}| = \sqrt{9 + 4 + 1} = \sqrt{14} \] 4. Calculate \( \cos\theta \): \[ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-5}{\sqrt{14} \sqrt{14}} \] \[ \cos\theta = -\frac{5}{14} \] 5. Find the angle \( \theta \): \[ \theta = \cos^{-1}\left(-\frac{5}{14}\right) \] Step 4: Final Answer:
The angle between the two vectors is \( \cos^{-1}\left(-\frac{5}{14}\right) \).