Find the shortest distance between the lines whose vector equations are :
`vec(r) = (hati + 2 hatj + 3 hatk ) + lambda (hati -3 hatj + 2 hatk) ` and
`vec(r) = 4 hati + 5 hatj + 6 hatk + mu (2 hati + 3 hatj + hatk)`.

Answer

To find the shortest distance between the two lines given by their vector equations, we can use the formula for the distance between two skew lines. The lines are represented in the following forms:

1. Line 1: \(\vec{r} = \vec{a} + \lambda \vec{b}\)
2. Line 2: \(\vec{r} = \vec{c} + \mu \vec{d}\)

### Step 1: Identify the vectors

From the given equations, we can identify the vectors:

- For Line 1:
- \(\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}\)
- \(\vec{b} = \hat{i} - 3\hat{j} + 2\hat{k}\)

- For Line 2:
- \(\vec{c} = 4\hat{i} + 5\hat{j} + 6\hat{k}\)
- \(\vec{d} = 2\hat{i} + 3\hat{j} + \hat{k}\)

### Step 2: Calculate \(\vec{c} - \vec{a}\)

Now, we calculate the vector \(\vec{c} - \vec{a}\):

\[ \vec{c} - \vec{a} = (4\hat{i} + 5\hat{j} + 6\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = (4 - 1)\hat{i} + (5 - 2)\hat{j} + (6 - 3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k} \]

### Step 3: Calculate \(\vec{b} \times \vec{d}\)

Next, we need to find the cross product \(\vec{b} \times \vec{d}\):

\[ \vec{b} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}, \quad \vec{d} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} \]

Using the determinant method for the cross product:

\[ \vec{b} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} -3 & 2 \\ 3 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -3 \\ 2 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} -3 & 2 \\ 3 & 1 \end{vmatrix} = (-3)(1) - (2)(3) = -3 - 6 = -9\)
2. \(\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = (1)(1) - (2)(2) = 1 - 4 = -3\)
3. \(\begin{vmatrix} 1 & -3 \\ 2 & 3 \end{vmatrix} = (1)(3) - (-3)(2) = 3 + 6 = 9\)

Putting it all together:

\[ \vec{b} \times \vec{d} = -9\hat{i} + 3\hat{j} + 9\hat{k} \]

### Step 4: Calculate the magnitude of \(\vec{b} \times \vec{d}\)

Now we find the magnitude of \(\vec{b} \times \vec{d}\):

\[ |\vec{b} \times \vec{d}| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19} \]

### Step 5: Calculate the shortest distance \(d\)

Finally, we can use the formula for the shortest distance \(d\) between the two lines:

\[ d = \frac{|(\vec{c} - \vec{a}) \cdot (\vec{b} \times \vec{d})|}{|\vec{b} \times \vec{d}|} \]

Calculating the dot product \((\vec{c} - \vec{a}) \cdot (\vec{b} \times \vec{d})\):

\[ \vec{c} - \vec{a} = 3\hat{i} + 3\hat{j} + 3\hat{k} \]
\[ \vec{b} \times \vec{d} = -9\hat{i} + 3\hat{j} + 9\hat{k} \]

Calculating the dot product:

\[ (3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-9\hat{i} + 3\hat{j} + 9\hat{k}) = 3(-9) + 3(3) + 3(9) = -27 + 9 + 27 = 9 \]

Now substituting back into the distance formula:

\[ d = \frac{|9|}{3\sqrt{19}} = \frac{9}{3\sqrt{19}} = \frac{3}{\sqrt{19}} \]

### Final Answer

The shortest distance between the two lines is:

\[ d = \frac{3}{\sqrt{19}} \]

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