To find the angle between the vectors \( \mathbf{a} = (1, 1, -1) \) and \( \mathbf{b} = (1, 0, 0) \), we can use the dot product formula, which relates the cosine of the angle between two vectors to their dot product and magnitudes:
\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \]
- Calculate the dot product \( \mathbf{a} \cdot \mathbf{b} \):
\[ \mathbf{a} \cdot \mathbf{b} = (1)(1) + (1)(0) + (-1)(0) = 1 + 0 + 0 = 1 \]
- Calculate the magnitudes \( |\mathbf{a}| \) and \( |\mathbf{b}| \):
\[ |\mathbf{a}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]
\[ |\mathbf{b}| = \sqrt{1^2 + 0^2 + 0^2} = \sqrt{1} = 1 \]
- Substitute into the dot product formula:
\[ 1 = \sqrt{3} \cdot 1 \cdot \cos \theta \]
\[ \cos \theta = \frac{1}{\sqrt{3}} \]
- Find the angle \( \theta \):
To find \( \theta \), we can take the arccosine:
\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \]
This value can also be expressed in degrees if necessary. Calculating \( \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \) gives:
\[ \theta \approx 54.74^\circ \]
Thus, the angle between the vectors \( \mathbf{a} \) and \( \mathbf{b} \) is approximately \( 54.74^\circ \).
1 answer