The position vectors of points A,B,C and D are `A=3hati+4hatj+5hatk`,`B=4hati+5hatj+6hatk`,`C=7hati+9hatj+3hatk`, and `D=4hati+6hatj`, then the displacement vectors AB and CD are

Options

A. perpendicular
B. parallel
C. antiparallel
D. inclined at an angle of `60^@`

Answer

To solve the problem, we need to find the displacement vectors \( \vec{AB} \) and \( \vec{CD} \) and then determine the angle between them.

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

The displacement vector \( \vec{AB} \) is given by:

\[ \vec{AB} = \vec{B} - \vec{A} \]

Given 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 substituting the values:

\[ \vec{AB} = (4\hat{i} + 5\hat{j} + 6\hat{k}) - (3\hat{i} + 4\hat{j} + 5\hat{k}) \]

Calculating component-wise:

\[ \vec{AB} = (4 - 3)\hat{i} + (5 - 4)\hat{j} + (6 - 5)\hat{k} = 1\hat{i} + 1\hat{j} + 1\hat{k} \]

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:

\[ \vec{CD} = \vec{D} - \vec{C} \]

Given the position vectors:

\[ \vec{C} = 7\hat{i} + 9\hat{j} + 3\hat{k} \]
\[ \vec{D} = 4\hat{i} + 6\hat{j} \]

Now substituting the values:

\[ \vec{CD} = (4\hat{i} + 6\hat{j}) - (7\hat{i} + 9\hat{j} + 3\hat{k}) \]

Calculating component-wise:

\[ \vec{CD} = (4 - 7)\hat{i} + (6 - 9)\hat{j} + (0 - 3)\hat{k} = -3\hat{i} - 3\hat{j} - 3\hat{k} \]

Thus,

\[ \vec{CD} = -3\hat{i} - 3\hat{j} - 3\hat{k} \]

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

To find the angle \( \theta \) between the vectors, we use the formula:

\[ \cos \theta = \frac{\vec{AB} \cdot \vec{CD}}{|\vec{AB}| |\vec{CD}|} \]

First, we calculate the dot product \( \vec{AB} \cdot \vec{CD} \):

\[ \vec{AB} \cdot \vec{CD} = (1)(-3) + (1)(-3) + (1)(-3) = -3 - 3 - 3 = -9 \]

Next, we find 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 back into the cosine formula:

\[ \cos \theta = \frac{-9}{\sqrt{3} \cdot 3\sqrt{3}} = \frac{-9}{9} = -1 \]

### Step 4: Determine the angle \( \theta \)

Since \( \cos \theta = -1 \), we have:

\[ \theta = 180^\circ \]

### Conclusion

The vectors \( \vec{AB} \) and \( \vec{CD} \) are anti-parallel since they are in opposite directions.