Let \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) be the unit vectors in the x, y and z directions, respectively. If the vector \(\hat{i} + \hat{j}\) is rotated about positive \(\hat{k}\) by 135°, one gets
GATE AE - 2022,
Answer: D
The result is \(-\sqrt{2}\hat{i}\).
The vector \(\hat{i} + \hat{j}\) has length \(\sqrt{2}\) and points at 45 degrees from the x axis in the x-y plane. A rotation about \(+\hat{k}\) is counterclockwise seen from above and leaves the length unchanged, so the vector ends at 45 + 135 = 180 degrees:
\[ \begin{aligned} &\sqrt{2}(\cos 180^{\circ}\,\hat{i} + \sin 180^{\circ}\,\hat{j}) \\ &= -\sqrt{2}\hat{i} \end{aligned} \]
\(-\hat{i}\) has the right direction but the wrong length, since a rotation cannot shrink a vector. \(-\hat{j}\) and \(-\frac{1}{\sqrt{2}}\hat{j}\) point at 270 degrees, which a 225-degree rotation would give.
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