Answer :
To find the angle between the vectors [tex]\(\vec{A} = \vec{i} + 2\vec{j} - \vec{k}\)[/tex] and [tex]\(\vec{B} = \vec{i} + \vec{j} - 2\vec{k}\)[/tex], we need to use the dot product formula and the magnitudes of the vectors.
Let's proceed step-by-step:
1. Calculate the dot product [tex]\(\vec{A} \cdot \vec{B}\)[/tex]:
The dot product of two vectors [tex]\(\vec{A} = \langle a_1, a_2, a_3 \rangle\)[/tex] and [tex]\(\vec{B} = \langle b_1, b_2, b_3 \rangle\)[/tex] is given by:
[tex]\[ \vec{A} \cdot \vec{B} = a_1 b_1 + a_2 b_2 + a_3 b_3 \][/tex]
For [tex]\(\vec{A} = \langle 1, 2, -1 \rangle\)[/tex] and [tex]\(\vec{B} = \langle 1, 1, -2 \rangle\)[/tex]:
[tex]\[ \vec{A} \cdot \vec{B} = (1 \cdot 1) + (2 \cdot 1) + (-1 \cdot -2) = 1 + 2 + 2 = 5 \][/tex]
2. Calculate the magnitudes of [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex]:
The magnitude of a vector [tex]\(\vec{A} = \langle a_1, a_2, a_3 \rangle\)[/tex] is given by:
[tex]\[ \|\vec{A}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \][/tex]
For [tex]\(\vec{A} = \langle 1, 2, -1 \rangle\)[/tex]:
[tex]\[ \|\vec{A}\| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \approx 2.449 \][/tex]
For [tex]\(\vec{B} = \langle 1, 1, -2 \rangle\)[/tex]:
[tex]\[ \|\vec{B}\| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \approx 2.449 \][/tex]
3. Calculate the cosine of the angle between [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex]:
The cosine of the angle [tex]\(\theta\)[/tex] between two vectors can be found using the dot product and magnitudes:
[tex]\[ \cos(\theta) = \frac{\vec{A} \cdot \vec{B}}{\|\vec{A}\| \|\vec{B}\|} \][/tex]
Substituting the values:
[tex]\[ \cos(\theta) = \frac{5}{(2.449)(2.449)} \approx 0.833 \][/tex]
4. Calculate the angle [tex]\(\theta\)[/tex]:
Finally, to find the angle [tex]\(\theta\)[/tex], we use the inverse cosine function:
[tex]\[ \theta = \cos^{-1}(0.833) \approx 0.586 \text{ radians} \][/tex]
Converting radians to degrees:
[tex]\[ \theta \approx 0.586 \times \frac{180}{\pi} \approx 33.56^\circ \][/tex]
Thus, the angle between [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] is approximately [tex]\(33.56^\circ\)[/tex].
Therefore, the correct answer from the given choices is:
[tex]\[ (3) \, 30^\circ \][/tex]
Let's proceed step-by-step:
1. Calculate the dot product [tex]\(\vec{A} \cdot \vec{B}\)[/tex]:
The dot product of two vectors [tex]\(\vec{A} = \langle a_1, a_2, a_3 \rangle\)[/tex] and [tex]\(\vec{B} = \langle b_1, b_2, b_3 \rangle\)[/tex] is given by:
[tex]\[ \vec{A} \cdot \vec{B} = a_1 b_1 + a_2 b_2 + a_3 b_3 \][/tex]
For [tex]\(\vec{A} = \langle 1, 2, -1 \rangle\)[/tex] and [tex]\(\vec{B} = \langle 1, 1, -2 \rangle\)[/tex]:
[tex]\[ \vec{A} \cdot \vec{B} = (1 \cdot 1) + (2 \cdot 1) + (-1 \cdot -2) = 1 + 2 + 2 = 5 \][/tex]
2. Calculate the magnitudes of [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex]:
The magnitude of a vector [tex]\(\vec{A} = \langle a_1, a_2, a_3 \rangle\)[/tex] is given by:
[tex]\[ \|\vec{A}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \][/tex]
For [tex]\(\vec{A} = \langle 1, 2, -1 \rangle\)[/tex]:
[tex]\[ \|\vec{A}\| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \approx 2.449 \][/tex]
For [tex]\(\vec{B} = \langle 1, 1, -2 \rangle\)[/tex]:
[tex]\[ \|\vec{B}\| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \approx 2.449 \][/tex]
3. Calculate the cosine of the angle between [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex]:
The cosine of the angle [tex]\(\theta\)[/tex] between two vectors can be found using the dot product and magnitudes:
[tex]\[ \cos(\theta) = \frac{\vec{A} \cdot \vec{B}}{\|\vec{A}\| \|\vec{B}\|} \][/tex]
Substituting the values:
[tex]\[ \cos(\theta) = \frac{5}{(2.449)(2.449)} \approx 0.833 \][/tex]
4. Calculate the angle [tex]\(\theta\)[/tex]:
Finally, to find the angle [tex]\(\theta\)[/tex], we use the inverse cosine function:
[tex]\[ \theta = \cos^{-1}(0.833) \approx 0.586 \text{ radians} \][/tex]
Converting radians to degrees:
[tex]\[ \theta \approx 0.586 \times \frac{180}{\pi} \approx 33.56^\circ \][/tex]
Thus, the angle between [tex]\(\vec{A}\)[/tex] and [tex]\(\vec{B}\)[/tex] is approximately [tex]\(33.56^\circ\)[/tex].
Therefore, the correct answer from the given choices is:
[tex]\[ (3) \, 30^\circ \][/tex]