Let $\vec{f} (t) = [t] \hat{i} + (t - [t]) \hat{j} + [t + 1] \hat{k}$ where [.] denotes the greatest integer function. Then the vectors $\vec{f} (\frac{5}{4}) and \vec{f} (t), 0 < t < 1$ is

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parallel to each other

perpendicular to each other

inclined at $\cos^{- 1} \frac{2}{\sqrt{7 (1 - t^{2})}}$

inclined at $\cos^{- 1} \frac{8 + t}{9 \sqrt{1 + t^{2}}}$

answer is D.

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Detailed Solution

$\begin{array}{l} \vec{f} (\frac{5}{4}) = [\frac{5}{4}] \hat{i} + (\frac{5}{4} - [\frac{5}{4}]) \hat{j} + [\frac{5}{4} + 1] \hat{k} \\ = \hat{i} + (\frac{5}{4} - 1) \hat{j} + 2 \hat{k} \\ = \hat{i} + \frac{1}{4} \hat{j} + 2 \hat{k} \end{array}$

$\begin{array}{l} when 0 < t < 1, \vec{f} (t) = 0 \vec{i} + {t - 0} \vec{j} + \vec{k} \\ = t \vec{j} + \vec{k} \\ so, \cos θ = \frac{2 + \frac{t}{4}}{|\vec{i} + \frac{1}{4} \vec{j} + 2 \vec{k}| | t \vec{j} + \vec{k} |} \\ \vec{f} (\frac{5}{4}) \cdot \vec{f} (t) = 2 + \frac{t}{4} \\ = \frac{2 + \frac{t}{4}}{\sqrt{1 + \frac{1}{16} + 4} \sqrt{1 + t^{2}}} = \frac{8 + t}{9 \sqrt{1 + t^{2}}} \end{array}$