The position vectors of points A, B, C and D are :
`vec(A) = 3hat(i) + 4hat(j) + 5hat(k), vec(B) = 4hat(i) + 5hat(j) + 6hat(k)`
`vec(C ) = 7hat(i) + 9hat(j) + 3hat(k)` and `vec(D) = 4hat(i) + 6hat(j)`
Then the displacement vectors `vec(AB)` and `vec(CD)` are :

Options

A. perpendicular
B. parallel
C. anti-parallel
D. inclined at an angle of `60^(@)`

Answer

To find the displacement vectors \(\vec{AB}\) and \(\vec{CD}\), we will follow these steps:

### Step 1: Calculate the displacement vector \(\vec{AB}\)

The displacement vector \(\vec{AB}\) is given by:
\[ \vec{AB} = \vec{B} - \vec{A} \]

Substituting the position vectors:
\[ \vec{A} = 3\hat{i} + 4\hat{j} + 5\hat{k} \]
\[ \vec{B} = 4\hat{i} + 5\hat{j} + 6\hat{k} \]

Now, performing the subtraction:
\[ \vec{AB} = (4\hat{i} + 5\hat{j} + 6\hat{k}) - (3\hat{i} + 4\hat{j} + 5\hat{k}) \]
\[ = (4 - 3)\hat{i} + (5 - 4)\hat{j} + (6 - 5)\hat{k} \]
\[ = 1\hat{i} + 1\hat{j} + 1\hat{k} \]
Thus,
\[ \vec{AB} = \hat{i} + \hat{j} + \hat{k} \]

### Step 2: Calculate the displacement vector \(\vec{CD}\)

The displacement vector \(\vec{CD}\) is given by:
\[ \vec{CD} = \vec{D} - \vec{C} \]

Substituting the position vectors:
\[ \vec{C} = 7\hat{i} + 9\hat{j} + 3\hat{k} \]
\[ \vec{D} = 4\hat{i} + 6\hat{j} \]

Now, performing the subtraction:
\[ \vec{CD} = (4\hat{i} + 6\hat{j}) - (7\hat{i} + 9\hat{j} + 3\hat{k}) \]
\[ = (4 - 7)\hat{i} + (6 - 9)\hat{j} + (0 - 3)\hat{k} \]
\[ = -3\hat{i} - 3\hat{j} - 3\hat{k} \]
Thus,
\[ \vec{CD} = -3\hat{i} - 3\hat{j} - 3\hat{k} \]

### Step 3: Determine the angle between \(\vec{AB}\) and \(\vec{CD}\)

To find the angle \(\theta\) between the vectors \(\vec{AB}\) and \(\vec{CD}\), we use the dot product formula:
\[ \cos(\theta) = \frac{\vec{AB} \cdot \vec{CD}}{|\vec{AB}| |\vec{CD}|} \]

First, we calculate the dot product \(\vec{AB} \cdot \vec{CD}\):
\[ \vec{AB} \cdot \vec{CD} = (1\hat{i} + 1\hat{j} + 1\hat{k}) \cdot (-3\hat{i} - 3\hat{j} - 3\hat{k}) \]
\[ = (1)(-3) + (1)(-3) + (1)(-3) = -3 - 3 - 3 = -9 \]

Next, we calculate the magnitudes of \(\vec{AB}\) and \(\vec{CD}\):
\[ |\vec{AB}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]
\[ |\vec{CD}| = \sqrt{(-3)^2 + (-3)^2 + (-3)^2} = \sqrt{27} = 3\sqrt{3} \]

Now substituting these into the cosine formula:
\[ \cos(\theta) = \frac{-9}{\sqrt{3} \cdot 3\sqrt{3}} = \frac{-9}{9} = -1 \]

This implies:
\[ \theta = 180^\circ \]

### Conclusion

Since the angle between \(\vec{AB}\) and \(\vec{CD}\) is \(180^\circ\), the vectors are anti-parallel.

### Final Answer:
\(\vec{AB} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{CD} = -3\hat{i} - 3\hat{j} - 3\hat{k}\) are anti-parallel.

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