Question

Let $\overrightarrow{f}\left(t\right)=\left[t\right]\overrightarrow{i}+\{t-\left[t\right]\}\overrightarrow{j}+[t+1]\overrightarrow{k}$, where $[\cdot ]$ denotes the greatest integer function, then the vectors $\overrightarrow{f}\left(\frac{5}{4}\right)$$\overrightarrow{f}\left(t\right),0<t<1$, are

• A. Perpendicular to each other
• B. Parallel to each other
• C. Inclined at an angle ${\cos }^{-1}\frac{2}{\sqrt{5(1+t^2)}}$
• D. Inclined at ${\cos }^{-1}\frac{8+t}{9\sqrt{1+t^2}}$

Answer 1

The correct option is D Inclined at ${\cos }^{-1}\frac{8+t}{9\sqrt{1+t^2}}$

Note that $\overrightarrow{f}\left(\frac{5}{4}\right)=\overrightarrow{i}+\frac{1}{4}\overrightarrow{j}+2\overrightarrow{k}$.
For $0<t<1$, we have $\left[t\right]=0,\left\{t\right\}=t,\left[t+1\right]=1$.
Hence $\overrightarrow{f}\left(t\right)=t\overrightarrow{j}+\overrightarrow{k}$
Therefore the angle between the vectors $\overrightarrow{f}\left(\frac{5}{4}\right)$ and $\overrightarrow{f}\left(t\right)$ is ${\cos }^{-1}\left(\frac{\overrightarrow{f}\left(\frac{5}{4}\right)\bullet \overrightarrow{f}\left(t\right)}{\left|\overrightarrow{f}\left(\frac{5}{4}\right)\right|\left|\overrightarrow{f}\left(t\right)\right|}\right)$
Now, $\overrightarrow{f}\left(\frac{5}{4}\right)\bullet \overrightarrow{f}\left(t\right)=2+\frac{t}{4}$.

Also, $\left|\overrightarrow{f}\left(\frac{5}{4}\right)\right|=\sqrt{\frac{81}{16}}=\frac{9}{4}$ and $\left|\overrightarrow{f}\left(t\right)\right|=\sqrt{t^2+1}$

Hence the angle is ${\cos }^{-1}\left(\frac{2+\frac{t}{4}}{\frac{9}{4}\sqrt{t^2+1}}\right)$

Simplifying we get, the angle between the two vectors is ${\cos }^{-1}\left(\frac{8+t}{9\sqrt{1+t^2}}\right)$