A line passing through the point A with position vector `vec(a) = 4 hat(i) + 2 hat(j) + 2 hat(k)` is parallel to the vector `vec(b) = 2 hat(i) + 3 hat(j) + 6 hat(k)`. Find the length of the perpendicular drawn on this line from a point P with position vector `vec(r_(1)) = hat(i) + 2 hat(j) + 3 hat(k)`.

Answer

To solve the problem step by step, we will follow the instructions provided in the video transcript and apply the necessary mathematical concepts related to 3D geometry.

### Step 1: Identify the position vectors
We have:
- Point A with position vector \( \vec{a} = 4\hat{i} + 2\hat{j} + 2\hat{k} \)
- Direction vector \( \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} \)
- Point P with position vector \( \vec{r_1} = \hat{i} + 2\hat{j} + 3\hat{k} \)

### Step 2: Write the equation of the line
The equation of the line passing through point A and parallel to vector \( \vec{b} \) can be expressed as:
\[ \vec{r} = \vec{a} + \lambda \vec{b} \]
Substituting the values:
\[ \vec{r} = (4\hat{i} + 2\hat{j} + 2\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \]
This gives us:
\[ \vec{r} = (4 + 2\lambda)\hat{i} + (2 + 3\lambda)\hat{j} + (2 + 6\lambda)\hat{k} \]

### Step 3: Find the vector \( \vec{r_1} - \vec{a} \)
Now, we need to calculate \( \vec{r_1} - \vec{a} \):
\[ \vec{r_1} - \vec{a} = (\hat{i} + 2\hat{j} + 3\hat{k}) - (4\hat{i} + 2\hat{j} + 2\hat{k}) \]
Calculating this gives:
\[ \vec{r_1} - \vec{a} = (-3\hat{i} + 0\hat{j} + 1\hat{k}) = -3\hat{i} + \hat{k} \]

### Step 4: Compute the cross product \( (\vec{r_1} - \vec{a}) \times \vec{b} \)
Next, we compute the cross product:
\[ \vec{u} = \vec{r_1} - \vec{a} = -3\hat{i} + 0\hat{j} + 1\hat{k} \]
\[ \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k} \]
The cross product \( \vec{u} \times \vec{b} \) is calculated as follows:
\[ \vec{u} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 0 & 1 \\ 2 & 3 & 6 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}(0 \cdot 6 - 1 \cdot 3) - \hat{j}(-3 \cdot 6 - 1 \cdot 2) + \hat{k}(-3 \cdot 3 - 0 \cdot 2) \]
\[ = \hat{i}(0 - 3) - \hat{j}(-18 - 2) + \hat{k}(-9) \]
\[ = -3\hat{i} + 20\hat{j} - 9\hat{k} \]

### Step 5: Find the modulus of the cross product
Now, we find the modulus of the cross product:
\[ |\vec{u} \times \vec{b}| = \sqrt{(-3)^2 + 20^2 + (-9)^2} = \sqrt{9 + 400 + 81} = \sqrt{490} \]

### Step 6: Find the modulus of vector \( \vec{b} \)
Next, we find the modulus of vector \( \vec{b} \):
\[ |\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

### Step 7: Calculate the perpendicular distance
Finally, the length of the perpendicular drawn from point P to the line is given by:
\[ d = \frac{|\vec{u} \times \vec{b}|}{|\vec{b}|} = \frac{\sqrt{490}}{7} \]
Simplifying this gives:
\[ d = \frac{\sqrt{490}}{7} = \frac{7\sqrt{10}}{7} = \sqrt{10} \]

### Final Answer
The length of the perpendicular drawn on this line from point P is \( \sqrt{10} \).

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