\( \begin{array}{l}\text { Let } x, y \text { be orhogonal vectors in a } \\ \text { vector space } \mathrm{V} \text { and }\end{array} \) \( \|x\|=\sqrt{5},\|y\|=\sqrt{7} \) Evaluate \( \|2 x-y\| \)

To find the norm \( \|2x - y\| \), we can use the properties of norms and the fact that \(x\) and \(y\) are orthogonal vectors. First, we calculate \( \|2x - y\|^2 \): \[ \|2x - y\|^2 = \|2x\|^2 + \|y\|^2 - 2 \langle 2x, y \rangle \] Since \(x\) and \(y\) are orthogonal, \( \langle x, y \rangle = 0 \), which means \( \langle 2x, y \rangle = 0 \) as well. Therefore, the formula simplifies to: \[ \|2x - y\|^2 = \|2x\|^2 + \|y\|^2 \] Next, we calculate \( \|2x\|^2 \): \[ \|2x\|^2 = (2\|x\|)^2 = 4\|x\|^2 = 4(\sqrt{5})^2 = 4 \cdot 5 = 20 \] Now, calculate \( \|y\|^2 \): \[ \|y\|^2 = (\sqrt{7})^2 = 7 \] Thus, we can substitute these values back into our squared norm equation: \[ \|2x - y\|^2 = 20 + 7 = 27 \] Finally, to find \( \|2x - y\| \), we take the square root: \[ \|2x - y\| = \sqrt{27} = 3\sqrt{3} \] Therefore, the final answer is: \[ \|2x - y\| = 3\sqrt{3} \]