Question

Find the angle between a and b, where a = (1, 1, −1) and b = (1, 0, 0)

Answer 1

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:

\[
\mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos \theta
\]

1. **Calculate the dot product \( \mathbf{a} \cdot \mathbf{b} \)**:

\[
\mathbf{a} \cdot \mathbf{b} = (1)(1) + (1)(0) + (-1)(0) = 1 + 0 + 0 = 1
\]

2. **Calculate the magnitudes \( \|\mathbf{a}\| \) and \( \|\mathbf{b}\| \)**:

\[
\|\mathbf{a}\| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}
\]

\[
\|\mathbf{b}\| = \sqrt{1^2 + 0^2 + 0^2} = \sqrt{1} = 1
\]

3. **Substitute into the dot product formula**:

\[
1 = \sqrt{3} \cdot 1 \cdot \cos \theta
\]

\[
\cos \theta = \frac{1}{\sqrt{3}}
\]

4. **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 \( 54.74^\circ \).