The distance between (5, 1, 3) and the line x = 3, y = 7 + t, z = 1 + t is

Options

A. 4
B. 2
C. 6
D. 8

Answer

To find the distance between the point \( P(5, 1, 3) \) and the line given by the equations \( x = 3 \), \( y = 7 + t \), and \( z = 1 + t \), we can follow these steps:

### Step 1: Identify the point on the line and the direction vector
The line can be represented in vector form. The point on the line can be taken as \( A(3, 7, 1) \) when \( t = 0 \). The direction vector \( \mathbf{b} \) of the line can be derived from the equations:
- For \( y = 7 + t \), the coefficient of \( t \) is 1.
- For \( z = 1 + t \), the coefficient of \( t \) is also 1.
- For \( x = 3 \), there is no change in \( x \) (the coefficient is 0).

Thus, the direction vector \( \mathbf{b} = (0, 1, 1) \).

### Step 2: Find the vector from point \( P \) to point \( A \)
The vector \( \mathbf{PA} \) from point \( P(5, 1, 3) \) to point \( A(3, 7, 1) \) is calculated as follows:
\[ \mathbf{PA} = A - P = (3 - 5, 7 - 1, 1 - 3) = (-2, 6, -2) \]

### Step 3: Calculate the cross product \( \mathbf{b} \times \mathbf{PA} \)
To find the distance, we need the cross product of \( \mathbf{b} \) and \( \mathbf{PA} \):
\[ \mathbf{b} = (0, 1, 1), \quad \mathbf{PA} = (-2, 6, -2) \]
Using the determinant to calculate the cross product:
\[ \mathbf{b} \times \mathbf{PA} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 1 \\ -2 & 6 & -2 \end{vmatrix} \]
Calculating the determinant:
\[ = \mathbf{i}(1 \cdot (-2) - 1 \cdot 6) - \mathbf{j}(0 \cdot (-2) - 1 \cdot (-2)) + \mathbf{k}(0 \cdot 6 - 1 \cdot (-2)) \]
\[ = \mathbf{i}(-2 - 6) - \mathbf{j}(0 + 2) + \mathbf{k}(0 + 2) \]
\[ = -8\mathbf{i} - 2\mathbf{j} + 2\mathbf{k} \]
Thus, \( \mathbf{b} \times \mathbf{PA} = (-8, -2, 2) \).

### Step 4: Calculate the magnitudes
Now, we find the magnitudes of \( \mathbf{b} \) and \( \mathbf{b} \times \mathbf{PA} \):
\[ |\mathbf{b}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2} \]
\[ |\mathbf{b} \times \mathbf{PA}| = \sqrt{(-8)^2 + (-2)^2 + 2^2} = \sqrt{64 + 4 + 4} = \sqrt{72} = 6\sqrt{2} \]

### Step 5: Calculate the distance
The distance \( d \) from the point to the line is given by the formula:
\[ d = \frac{|\mathbf{b} \times \mathbf{PA}|}{|\mathbf{b}|} \]
Substituting the values we found:
\[ d = \frac{6\sqrt{2}}{\sqrt{2}} = 6 \]

### Final Answer
The distance between the point \( (5, 1, 3) \) and the line is \( 6 \) units.

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