If A, B, C, D are the points with position vectors : `hat(i)+hat(j)-hat(k), 2hat(i)-hat(j)+3hat(k), 2hat(i)-3hat(k), 3hat(i)-2hat(j)+hat(k)` respectively. Find the projection of `vec(AB)` along CD.

Answer

To find the projection of vector \( \vec{AB} \) along \( \vec{CD} \), we will follow these steps:

### Step 1: Find the position vectors of points A, B, C, and D.

Given:
- \( \vec{A} = \hat{i} + \hat{j} - \hat{k} \)
- \( \vec{B} = 2\hat{i} - \hat{j} + 3\hat{k} \)
- \( \vec{C} = 2\hat{i} - 3\hat{k} \)
- \( \vec{D} = 3\hat{i} - 2\hat{j} + \hat{k} \)

### Step 2: Calculate the vector \( \vec{AB} \).

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

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

Substituting the position vectors:

\[ \vec{AB} = (2\hat{i} - \hat{j} + 3\hat{k}) - (\hat{i} + \hat{j} - \hat{k}) \]

Now, simplify:

\[ \vec{AB} = (2\hat{i} - \hat{i}) + (-\hat{j} - \hat{j}) + (3\hat{k} + \hat{k}) = \hat{i} - 2\hat{j} + 4\hat{k} \]

### Step 3: Calculate the vector \( \vec{CD} \).

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

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

Substituting the position vectors:

\[ \vec{CD} = (3\hat{i} - 2\hat{j} + \hat{k}) - (2\hat{i} - 3\hat{k}) \]

Now, simplify:

\[ \vec{CD} = (3\hat{i} - 2\hat{i}) + (-2\hat{j}) + (\hat{k} + 3\hat{k}) = \hat{i} - 2\hat{j} + 4\hat{k} \]

### Step 4: Find the projection of \( \vec{AB} \) along \( \vec{CD} \).

The formula for the projection of vector \( \vec{u} \) along vector \( \vec{v} \) is given by:

\[ \text{proj}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{|\vec{v}|^2} \vec{v} \]

In our case, \( \vec{u} = \vec{AB} \) and \( \vec{v} = \vec{CD} \).

### Step 5: Calculate the dot product \( \vec{AB} \cdot \vec{CD} \).

\[ \vec{AB} \cdot \vec{CD} = (\hat{i} - 2\hat{j} + 4\hat{k}) \cdot (\hat{i} - 2\hat{j} + 4\hat{k}) \]

Calculating the dot product:

\[ = (1)(1) + (-2)(-2) + (4)(4) = 1 + 4 + 16 = 21 \]

### Step 6: Calculate the magnitude of \( \vec{CD} \).

\[ |\vec{CD}| = \sqrt{(1)^2 + (-2)^2 + (4)^2} = \sqrt{1 + 4 + 16} = \sqrt{21} \]

### Step 7: Calculate the projection.

Now, substitute into the projection formula:

\[ \text{proj}_{\vec{CD}} \vec{AB} = \frac{21}{|\vec{CD}|^2} \vec{CD} \]

Since \( |\vec{CD}|^2 = 21 \):

\[ \text{proj}_{\vec{CD}} \vec{AB} = \frac{21}{21} \vec{CD} = \vec{CD} \]

Thus, the projection of \( \vec{AB} \) along \( \vec{CD} \) is:

\[ \text{proj}_{\vec{CD}} \vec{AB} = \hat{i} - 2\hat{j} + 4\hat{k} \]

### Final Answer:

The projection of \( \vec{AB} \) along \( \vec{CD} \) is \( \hat{i} - 2\hat{j} + 4\hat{k} \).

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