A line passes through two points `A(2,-3,-1)` and `B(8,-1,2)`. The coordinates of a point on this lie at distance of 14 units from a are

Options

A. `(14,1,5)`
B. `(-10,-7,7)`
C. `(86,25,41)`
D. none of these

Explanation

To find the coordinates of a point on the line passing through points A(2, -3, -1) and B(8, -1, 2) that is at a distance of 14 units from point A, we can follow these steps:

### Step 1: Find the direction vector of the line AB
The direction vector \( \vec{AB} \) can be calculated as:
\[ \vec{AB} = \vec{B} - \vec{A} = (8 - 2, -1 - (-3), 2 - (-1)) = (6, 2, 3) \]

### Step 2: Write the parametric equations of the line
Using point A as the reference point and the direction vector \( \vec{AB} \), we can express the coordinates of any point \( P \) on the line in terms of a parameter \( \lambda \):
\[ x = 2 + 6\lambda \]
\[ y = -3 + 2\lambda \]
\[ z = -1 + 3\lambda \]

### Step 3: Use the distance formula
The distance \( d \) between point A and point P is given by:
\[ d = \sqrt{(x - 2)^2 + (y + 3)^2 + (z + 1)^2} \]
We set this equal to 14:
\[ \sqrt{(6\lambda)^2 + (2\lambda)^2 + (3\lambda)^2} = 14 \]

### Step 4: Simplify the equation
Squaring both sides, we have:
\[ (6\lambda)^2 + (2\lambda)^2 + (3\lambda)^2 = 14^2 \]
\[ 36\lambda^2 + 4\lambda^2 + 9\lambda^2 = 196 \]
\[ 49\lambda^2 = 196 \]

### Step 5: Solve for \( \lambda \)
Dividing both sides by 49:
\[ \lambda^2 = 4 \implies \lambda = 2 \text{ or } \lambda = -2 \]

### Step 6: Find the coordinates of point P for both values of \( \lambda \)
1. For \( \lambda = 2 \):
\[ x = 2 + 6(2) = 14 \]
\[ y = -3 + 2(2) = 1 \]
\[ z = -1 + 3(2) = 5 \]
Thus, one point \( P_1 \) is \( (14, 1, 5) \).

2. For \( \lambda = -2 \):
\[ x = 2 + 6(-2) = -10 \]
\[ y = -3 + 2(-2) = -7 \]
\[ z = -1 + 3(-2) = -7 \]
Thus, another point \( P_2 \) is \( (-10, -7, -7) \).

### Final Answer
The coordinates of the points on the line that are at a distance of 14 units from point A are \( (14, 1, 5) \) and \( (-10, -7, -7) \).

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