Let ` vecf(t)=[t] hat i+(t-[t]) hat j+[t+1] hat k , w h e r e[dot]` denotes the greatest integer function. Then the vectors `vecf(5/4)` and `vecf(t)` ,`0 < t < 1` are `(a)`parallel to each other `(b)`perpendicular to each other `(c)`inclined at `cos^(-1)(2/(sqrt(7(1-t^2))))` `(d)`inclined at `cos^(-1)((8+t)/(9*sqrt(1+t^2)))`

Options

A. parallel to each other
B. perpendicular to each other
C. inclined at `cos^(-1) 2/(sqrt7(1-t^(2))`
D. inclined at `cos ^(-1) (8+t)/(9sqrt(1+t^(2)))`

Explanation

To solve the problem, we need to analyze the vectors given in the question and determine the relationship between `vecf(5/4)` and `vecf(t)` for `0 < t < 1`.

### Step 1: Calculate `vecf(5/4)`

Given the function:
\[ \vecf(t) = [t] \hat{i} + (t - [t]) \hat{j} + [t + 1] \hat{k} \]
where `[t]` denotes the greatest integer function.

For \( t = \frac{5}{4} \):
- \([t] = [\frac{5}{4}] = 1\)
- \(t - [t] = \frac{5}{4} - 1 = \frac{1}{4}\)
- \([t + 1] = [\frac{5}{4} + 1] = [\frac{9}{4}] = 2\)

Thus, we have:
\[ \vecf\left(\frac{5}{4}\right) = 1 \hat{i} + \frac{1}{4} \hat{j} + 2 \hat{k} = \hat{i} + \frac{1}{4} \hat{j} + 2 \hat{k} \]

### Step 2: Calculate `vecf(t)` for `0 < t < 1`

For \( 0 < t < 1 \):
- \([t] = 0\)
- \(t - [t] = t - 0 = t\)
- \([t + 1] = [t + 1] = 1\)

Thus, we have:
\[ \vecf(t) = 0 \hat{i} + t \hat{j} + 1 \hat{k} = t \hat{j} + \hat{k} \]

### Step 3: Find the dot product of `vecf(5/4)` and `vecf(t)`

Now we need to find the dot product:
\[ \vecf\left(\frac{5}{4}\right) \cdot \vecf(t) = \left(1 \hat{i} + \frac{1}{4} \hat{j} + 2 \hat{k}\right) \cdot \left(0 \hat{i} + t \hat{j} + 1 \hat{k}\right) \]
Calculating the dot product:
\[ = 1 \cdot 0 + \frac{1}{4} \cdot t + 2 \cdot 1 = \frac{t}{4} + 2 \]

### Step 4: Calculate the magnitudes of the vectors

Magnitude of \(\vecf\left(\frac{5}{4}\right)\):
\[ \|\vecf\left(\frac{5}{4}\right)\| = \sqrt{1^2 + \left(\frac{1}{4}\right)^2 + 2^2} = \sqrt{1 + \frac{1}{16} + 4} = \sqrt{\frac{16 + 1 + 64}{16}} = \sqrt{\frac{81}{16}} = \frac{9}{4} \]

Magnitude of \(\vecf(t)\):
\[ \|\vecf(t)\| = \sqrt{0^2 + t^2 + 1^2} = \sqrt{t^2 + 1} \]

### Step 5: Calculate the cosine of the angle between the vectors

Using the formula for the cosine of the angle:
\[ \cos \theta = \frac{\vecf\left(\frac{5}{4}\right) \cdot \vecf(t)}{\|\vecf\left(\frac{5}{4}\right)\| \|\vecf(t)\|} \]
Substituting the values:
\[ \cos \theta = \frac{\frac{t}{4} + 2}{\frac{9}{4} \cdot \sqrt{t^2 + 1}} \]
Simplifying:
\[ = \frac{t + 8}{9 \sqrt{t^2 + 1}} \]

### Step 6: Find the angle

Thus, the angle between the vectors is:
\[ \theta = \cos^{-1}\left(\frac{t + 8}{9 \sqrt{t^2 + 1}}\right) \]

### Conclusion

After checking the options, we find that the correct answer corresponds to option (d):
\[ \text{(d) inclined at } \cos^{-1}\left(\frac{8 + t}{9 \sqrt{1 + t^2}}\right) \]