Question

Angle between vectors \( \vec{a} = \hat{i} + \hat{j} - \hat{k} \) and \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \) is 60°.

Hint:

To calculate the angle between two vectors, use the dot product formula and the magnitudes of the vectors.

Answer 1

Step 1: Formula for angle between two vectors.
The angle \( \theta \) between two vectors \( \vec{a} \) and \( \vec{b} \) is given by the formula: \[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}. \] where \( \vec{a} \cdot \vec{b} \) is the dot product of the vectors and \( |\vec{a}| \) and \( |\vec{b}| \) are the magnitudes of the vectors.

Step 2: Calculating the dot product.
The dot product \( \vec{a} \cdot \vec{b} \) is: \[ \vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1. \]

Step 3: Calculating the magnitudes.
The magnitudes of \( \vec{a} \) and \( \vec{b} \) are: \[ |\vec{a}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}, |\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}. \]

Step 4: Finding the angle.
Now, using the formula for the angle: \[ \cos \theta = \frac{-1}{\sqrt{3} \times \sqrt{3}} = \frac{-1}{3}, \theta = \cos^{-1}\left(\frac{-1}{3}\right) \approx 60^\circ. \]

Step 5: Conclusion.
Thus, the angle between the vectors is 60°, which makes the statement true.