To determine which vectors could be representations of \( w \), given that the magnitude of the vector \(-3w\) is 15, we can follow these steps:
1. Understanding the magnitude relationship: The magnitude of a vector is given by the formula \( \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2} \), where \(\mathbf{v} = \langle v_1, v_2 \rangle\).
2. Given information: We are told that \(\|-3w\| = 15\). We know from vector operations that the magnitude of \(-3w\) can be expressed as \(\|-3 w\| = 3 \|w\|\), due to the property that multiplying a vector by a scalar scales its magnitude by the absolute value of that scalar. Therefore:
[tex]\[
\|-3w\| = 3 \|w\| \Rightarrow 3 \|w\| = 15 \Rightarrow \|w\| = \frac{15}{3} = 5
\][/tex]
3. Magnitude calculation for possible vectors: Now, we need to determine which of the given vectors have a magnitude of 5.
- For the vector \( \langle 1, -9 \rangle \):
[tex]\[
\sqrt{1^2 + (-9)^2} = \sqrt{1 + 81} = \sqrt{82} \approx 9.06 \neq 5
\][/tex]
- For the vector \( \langle -3, 4 \rangle \):
[tex]\[
\sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\][/tex]
- For the vector \( \langle 4, 5 \rangle \):
[tex]\[
\sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \approx 6.4 \neq 5
\][/tex]
- For the vector \( \langle -5, -3 \rangle \):
[tex]\[
\sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.83 \neq 5
\][/tex]
- For the vector \( \langle 0, -5 \rangle \):
[tex]\[
\sqrt{0^2 + (-5)^2} = \sqrt{0 + 25} = \sqrt{25} = 5
\][/tex]
4. Conclusion: Therefore, the vectors that have the correct magnitude of 5 are:
[tex]\[
\langle -3, 4 \rangle \quad \text{and} \quad \langle 0, -5 \rangle
\][/tex]
Hence, the possible representations of the vector [tex]\( w \)[/tex] given the magnitude of [tex]\(-3w\)[/tex] are [tex]\(\langle -3, 4 \rangle\)[/tex] and [tex]\(\langle 0, -5 \rangle\)[/tex].
