The position vector of the foot of the perpendicular draw from the point `2hati-hatj+5hatk` to the line
`vecr=(11hati-2hatj-8hatk)+lamda(10hati-4hatj-11hatk)` is

Options

A. `hati+3hatj+2hatk`
B. `-hati+3hatj-2hatk`
C. `hati-3hatj-2hatk`
D. `hati+2hatj+3hatk`

Explanation

To find the position vector of the foot of the perpendicular drawn from the point \( \mathbf{P} = 2\hat{i} - \hat{j} + 5\hat{k} \) to the line given by the vector equation

\[ \mathbf{r} = (11\hat{i} - 2\hat{j} - 8\hat{k}) + \lambda(10\hat{i} - 4\hat{j} - 11\hat{k}), \]

we can follow these steps:

### Step 1: Identify the point and direction of the line

The line can be expressed in terms of a point \( \mathbf{A} \) and a direction vector \( \mathbf{d} \):
- Point \( \mathbf{A} = 11\hat{i} - 2\hat{j} - 8\hat{k} \)
- Direction vector \( \mathbf{d} = 10\hat{i} - 4\hat{j} - 11\hat{k} \)

### Step 2: Write the coordinates of point \( Q \) on the line

The coordinates of any point \( Q \) on the line can be expressed as:
\[ Q = (11 + 10\lambda)\hat{i} + (-2 - 4\lambda)\hat{j} + (-8 - 11\lambda)\hat{k} \]

### Step 3: Find the vector \( \mathbf{PQ} \)

The vector \( \mathbf{PQ} \) from point \( P \) to point \( Q \) can be calculated as:
\[ \mathbf{PQ} = Q - P = \left((11 + 10\lambda) - 2\right)\hat{i} + \left((-2 - 4\lambda) + 1\right)\hat{j} + \left((-8 - 11\lambda) - 5\right)\hat{k} \]
Simplifying this gives:
\[ \mathbf{PQ} = (10\lambda + 9)\hat{i} + (-4\lambda - 1)\hat{j} + (-11\lambda - 13)\hat{k} \]

### Step 4: Set up the dot product condition for perpendicularity

Since \( \mathbf{PQ} \) is perpendicular to the direction vector \( \mathbf{d} \), we have:
\[ \mathbf{PQ} \cdot \mathbf{d} = 0 \]
Calculating the dot product:
\[ (10\lambda + 9)(10) + (-4\lambda - 1)(-4) + (-11\lambda - 13)(-11) = 0 \]
Expanding this gives:
\[ 100\lambda + 90 + 16\lambda + 4 + 121\lambda + 143 = 0 \]
Combining like terms results in:
\[ (100 + 16 + 121)\lambda + (90 + 4 + 143) = 0 \]
\[ 237\lambda + 237 = 0 \]

### Step 5: Solve for \( \lambda \)

Setting the equation to zero:
\[ 237\lambda + 237 = 0 \implies \lambda = -1 \]

### Step 6: Substitute \( \lambda \) back to find \( Q \)

Substituting \( \lambda = -1 \) back into the coordinates of \( Q \):
\[ Q = (11 + 10(-1))\hat{i} + (-2 - 4(-1))\hat{j} + (-8 - 11(-1))\hat{k} \]
Calculating each component:
\[ Q = (11 - 10)\hat{i} + (-2 + 4)\hat{j} + (-8 + 11)\hat{k} \]
\[ Q = 1\hat{i} + 2\hat{j} + 3\hat{k} \]

### Final Answer

Thus, the position vector of the foot of the perpendicular is:
\[ \mathbf{Q} = \hat{i} + 2\hat{j} + 3\hat{k} \]