Which of the following is/are the points that is/are at a distance of 12 units from the point whose position vector is `(8hatk+10hatj-8hatk)` on the line which is parallel to `(2hati+hatj+2hatk)`?

Options

A. `16hati+14hatj`
B. `6hatj-16hatk`
C. `(16hati+18hatj-4hatk)`
D. none of these

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Identify the Given Information
The position vector of the point is given as:
\[ \vec{A} = 8\hat{i} + 10\hat{j} - 8\hat{k} \]
The direction vector of the line, which is parallel to the line, is:
\[ \vec{b} = 2\hat{i} + \hat{j} + 2\hat{k} \]
We need to find points that are at a distance of 12 units from point A.

### Step 2: Write the Equation of the Line
The general equation of a line in vector form can be written as:
\[ \vec{r} = \vec{A} + \lambda \vec{b} \]
Substituting the known values:
\[ \vec{r} = (8\hat{i} + 10\hat{j} - 8\hat{k}) + \lambda(2\hat{i} + \hat{j} + 2\hat{k}) \]

### Step 3: Simplify the Equation
Expanding the equation gives:
\[ \vec{r} = (8 + 2\lambda)\hat{i} + (10 + \lambda)\hat{j} + (-8 + 2\lambda)\hat{k} \]

### Step 4: Calculate the Distance
The distance from point A to point R (the point on the line) is given as 12 units. The distance formula in vector form is:
\[ |\vec{r} - \vec{A}| = 12 \]
Substituting \(\vec{r}\) and \(\vec{A}\):
\[ |(8 + 2\lambda - 8)\hat{i} + (10 + \lambda - 10)\hat{j} + (-8 + 2\lambda + 8)\hat{k}| = 12 \]
This simplifies to:
\[ |2\lambda \hat{i} + \lambda \hat{j} + 2\lambda \hat{k}| = 12 \]

### Step 5: Calculate the Magnitude
The magnitude can be calculated as:
\[ \sqrt{(2\lambda)^2 + (\lambda)^2 + (2\lambda)^2} = 12 \]
This simplifies to:
\[ \sqrt{4\lambda^2 + \lambda^2 + 4\lambda^2} = 12 \]
\[ \sqrt{9\lambda^2} = 12 \]
\[ 3|\lambda| = 12 \]
Thus,
\[ |\lambda| = 4 \implies \lambda = 4 \text{ or } \lambda = -4 \]

### Step 6: Find Points for \(\lambda = 4\) and \(\lambda = -4\)
1. For \(\lambda = 4\):
\[ \vec{r} = (8 + 2 \cdot 4)\hat{i} + (10 + 4)\hat{j} + (-8 + 2 \cdot 4)\hat{k} \]
\[ = (8 + 8)\hat{i} + (10 + 4)\hat{j} + (-8 + 8)\hat{k} \]
\[ = 16\hat{i} + 14\hat{j} + 0\hat{k} \]
So, the first point is \(16\hat{i} + 14\hat{j}\).

2. For \(\lambda = -4\):
\[ \vec{r} = (8 + 2 \cdot (-4))\hat{i} + (10 - 4)\hat{j} + (-8 + 2 \cdot (-4))\hat{k} \]
\[ = (8 - 8)\hat{i} + (10 - 4)\hat{j} + (-8 - 8)\hat{k} \]
\[ = 0\hat{i} + 6\hat{j} - 16\hat{k} \]
So, the second point is \(6\hat{j} - 16\hat{k}\).

### Step 7: Conclusion
The points that are at a distance of 12 units from the given point are:
1. \(16\hat{i} + 14\hat{j}\)
2. \(6\hat{j} - 16\hat{k}\)