Answer :
To solve the problem given vectors \( \mathbf{u} = \langle 5, -7 \rangle \) and \( \mathbf{v} = \langle -11, 3 \rangle \), we need to find \( 2\mathbf{v} - 6\mathbf{u} \) and then determine the magnitude (or norm) of that resulting vector. Here is a detailed step-by-step solution:
1. Compute \( 2\mathbf{v} \):
- Multiply each component of vector \(\mathbf{v}\) by 2.
- [tex]\[ 2\mathbf{v} = 2 \langle -11, 3 \rangle = \langle 2 \cdot -11, 2 \cdot 3 \rangle = \langle -22, 6 \rangle \][/tex]
2. Compute \( 6\mathbf{u} \):
- Multiply each component of vector \(\mathbf{u}\) by 6.
- [tex]\[ 6\mathbf{u} = 6 \langle 5, -7 \rangle = \langle 6 \cdot 5, 6 \cdot -7 \rangle = \langle 30, -42 \rangle \][/tex]
3. Compute \( 2\mathbf{v} - 6\mathbf{u} \):
- Subtract each corresponding component of \(6\mathbf{u}\) from \(2\mathbf{v}\).
- [tex]\[ 2\mathbf{v} - 6\mathbf{u} = \langle -22, 6 \rangle - \langle 30, -42 \rangle = \langle -22 - 30, 6 - (-42) \rangle = \langle -52, 48 \rangle \][/tex]
4. Calculate the magnitude (norm) of the resulting vector \( \langle -52, 48 \rangle \):
- The magnitude \( \|\mathbf{a}\| \) of vector \( \mathbf{a} = \langle a_1, a_2 \rangle \) is given by \( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2} \).
- [tex]\[ \| 2\mathbf{v} - 6\mathbf{u} \| = \| \langle -52, 48 \rangle \| = \sqrt{(-52)^2 + 48^2} = \sqrt{2704 + 2304} = \sqrt{5008} \approx 70.77 \][/tex]
Hence, the correct answers to fill in the blanks are:
- \( 2\mathbf{v} - 6\mathbf{u} = \langle -52, 48 \rangle \)
- \( \| 2\mathbf{v} - 6\mathbf{u} \| \approx 70.77 \)
Summarizing the filled blanks:
If vector \( u = \langle 5,-7 \rangle \) and \( v = \langle -11, 3 \rangle \):
- \( 2v - 6u = \langle -52, 48 \rangle \)
- [tex]\( \|2v - 6u\| \approx 70.77 \)[/tex]
1. Compute \( 2\mathbf{v} \):
- Multiply each component of vector \(\mathbf{v}\) by 2.
- [tex]\[ 2\mathbf{v} = 2 \langle -11, 3 \rangle = \langle 2 \cdot -11, 2 \cdot 3 \rangle = \langle -22, 6 \rangle \][/tex]
2. Compute \( 6\mathbf{u} \):
- Multiply each component of vector \(\mathbf{u}\) by 6.
- [tex]\[ 6\mathbf{u} = 6 \langle 5, -7 \rangle = \langle 6 \cdot 5, 6 \cdot -7 \rangle = \langle 30, -42 \rangle \][/tex]
3. Compute \( 2\mathbf{v} - 6\mathbf{u} \):
- Subtract each corresponding component of \(6\mathbf{u}\) from \(2\mathbf{v}\).
- [tex]\[ 2\mathbf{v} - 6\mathbf{u} = \langle -22, 6 \rangle - \langle 30, -42 \rangle = \langle -22 - 30, 6 - (-42) \rangle = \langle -52, 48 \rangle \][/tex]
4. Calculate the magnitude (norm) of the resulting vector \( \langle -52, 48 \rangle \):
- The magnitude \( \|\mathbf{a}\| \) of vector \( \mathbf{a} = \langle a_1, a_2 \rangle \) is given by \( \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2} \).
- [tex]\[ \| 2\mathbf{v} - 6\mathbf{u} \| = \| \langle -52, 48 \rangle \| = \sqrt{(-52)^2 + 48^2} = \sqrt{2704 + 2304} = \sqrt{5008} \approx 70.77 \][/tex]
Hence, the correct answers to fill in the blanks are:
- \( 2\mathbf{v} - 6\mathbf{u} = \langle -52, 48 \rangle \)
- \( \| 2\mathbf{v} - 6\mathbf{u} \| \approx 70.77 \)
Summarizing the filled blanks:
If vector \( u = \langle 5,-7 \rangle \) and \( v = \langle -11, 3 \rangle \):
- \( 2v - 6u = \langle -52, 48 \rangle \)
- [tex]\( \|2v - 6u\| \approx 70.77 \)[/tex]