Discussion
A vector is given as \(\vec{A} = 3\hat{j} - 4\hat{k}\). The vector parallel to \(\vec{A}\) and magnitude the same as that of the vector \(\hat{i} - 2\hat{j}\) is
Answer: D
The vector is \(\frac{1}{\sqrt{5}}(3\hat{j} - 4\hat{k})\).
The required vector is the unit vector along \(\vec{A}\) times the magnitude of \(\hat{i} - 2\hat{j}\).
\[ \begin{aligned} |\vec{A}| &= \sqrt{9 + 16} = 5, \quad \hat{A} = \tfrac{1}{5}(3\hat{j} - 4\hat{k}) \\ |\hat{i} - 2\hat{j}| &= \sqrt{5} \\ \text{vector} &= \tfrac{\sqrt{5}}{5}(3\hat{j} - 4\hat{k}) = \tfrac{1}{\sqrt{5}}(3\hat{j} - 4\hat{k}) \end{aligned} \]
The option \(\sqrt{5}(3\hat{j} - 4\hat{k})\) forgets to make \(\vec{A}\) a unit vector first, giving magnitude \(5\sqrt{5}\). The options containing \(\hat{i}\) are not parallel to \(\vec{A}\), which has no \(\hat{i}\) component.
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