Answer :
To find the angle between two vectors [tex]\(\mathbf{a} = i + 5j\)[/tex] and [tex]\(\mathbf{b} = 5i - j\)[/tex], we need to follow these steps:
1. Express the vectors in component form:
[tex]\[ \mathbf{a} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 5 \\ -1 \end{pmatrix} \][/tex]
2. Calculate the dot product of [tex]\(\mathbf{a}\)[/tex] and [tex]\(\mathbf{b}\)[/tex]:
[tex]\[ \mathbf{a} \cdot \mathbf{b} = (1)(5) + (5)(-1) = 5 - 5 = 0 \][/tex]
The dot product of [tex]\(\mathbf{a}\)[/tex] and [tex]\(\mathbf{b}\)[/tex] is 0.
3. Find the magnitudes of the vectors [tex]\(\mathbf{a}\)[/tex] and [tex]\(\mathbf{b}\)[/tex]:
[tex]\[ \| \mathbf{a} \| = \sqrt{1^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26} \approx 5.099 \][/tex]
[tex]\[ \| \mathbf{b} \| = \sqrt{5^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \approx 5.099 \][/tex]
The magnitudes of [tex]\(\mathbf{a}\)[/tex] and [tex]\(\mathbf{b}\)[/tex] are approximately 5.099 each.
4. Calculate the cosine of the angle [tex]\(\theta\)[/tex] between the two vectors using the dot product formula:
[tex]\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \| \| \mathbf{b} \|} \][/tex]
Since the dot product [tex]\(\mathbf{a} \cdot \mathbf{b} = 0\)[/tex], we have:
[tex]\[ \cos \theta = \frac{0}{5.099 \times 5.099} = 0 \][/tex]
5. Find the angle [tex]\(\theta\)[/tex] using the inverse cosine (arccos) function:
[tex]\[ \theta = \arccos(0) \][/tex]
The arccosine of 0 is [tex]\(\frac{\pi}{2}\)[/tex] radians (or 90 degrees).
Thus, the angle between the vectors [tex]\(\mathbf{a} = i + 5j\)[/tex] and [tex]\(\mathbf{b} = 5i - j\)[/tex] is 90 degrees.
1. Express the vectors in component form:
[tex]\[ \mathbf{a} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 5 \\ -1 \end{pmatrix} \][/tex]
2. Calculate the dot product of [tex]\(\mathbf{a}\)[/tex] and [tex]\(\mathbf{b}\)[/tex]:
[tex]\[ \mathbf{a} \cdot \mathbf{b} = (1)(5) + (5)(-1) = 5 - 5 = 0 \][/tex]
The dot product of [tex]\(\mathbf{a}\)[/tex] and [tex]\(\mathbf{b}\)[/tex] is 0.
3. Find the magnitudes of the vectors [tex]\(\mathbf{a}\)[/tex] and [tex]\(\mathbf{b}\)[/tex]:
[tex]\[ \| \mathbf{a} \| = \sqrt{1^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26} \approx 5.099 \][/tex]
[tex]\[ \| \mathbf{b} \| = \sqrt{5^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} \approx 5.099 \][/tex]
The magnitudes of [tex]\(\mathbf{a}\)[/tex] and [tex]\(\mathbf{b}\)[/tex] are approximately 5.099 each.
4. Calculate the cosine of the angle [tex]\(\theta\)[/tex] between the two vectors using the dot product formula:
[tex]\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \| \| \mathbf{b} \|} \][/tex]
Since the dot product [tex]\(\mathbf{a} \cdot \mathbf{b} = 0\)[/tex], we have:
[tex]\[ \cos \theta = \frac{0}{5.099 \times 5.099} = 0 \][/tex]
5. Find the angle [tex]\(\theta\)[/tex] using the inverse cosine (arccos) function:
[tex]\[ \theta = \arccos(0) \][/tex]
The arccosine of 0 is [tex]\(\frac{\pi}{2}\)[/tex] radians (or 90 degrees).
Thus, the angle between the vectors [tex]\(\mathbf{a} = i + 5j\)[/tex] and [tex]\(\mathbf{b} = 5i - j\)[/tex] is 90 degrees.
