The distance of the point (-2,4,-5) from the line `(x+3)/(3)=(y-4)/(5)=(z+8)/(6)` is

Options

A. `(sqrt(37))/(10)`
B. `sqrt((37)/(10))`
C. `(37)/(sqrt(10))`
D. `(37)/(10)`

Answer

To find the distance of the point \((-2, 4, -5)\) from the line given by the symmetric equations \(\frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6}\), we will follow these steps:

### Step 1: Identify a point on the line and the direction ratios
The line can be expressed in parametric form. Let \(k\) be the parameter. Then we can write:
- \(x = 3k - 3\)
- \(y = 5k + 4\)
- \(z = 6k - 8\)

From this, we can identify:
- A point on the line \(B\) when \(k = 0\):
\[ B(3(0) - 3, 5(0) + 4, 6(0) - 8) = (-3, 4, -8) \]
- The direction ratios of the line are \((3, 5, 6)\).

### Step 2: Find the vector from the point \(A\) to point \(B\)
Let \(A(-2, 4, -5)\) be the point from which we want to find the distance to the line. The vector \(AB\) is given by:
\[ AB = B - A = (-3 - (-2), 4 - 4, -8 - (-5)) = (-1, 0, -3) \]

### Step 3: Find the direction vector of the line
The direction vector of the line is:
\[ D = (3, 5, 6) \]

### Step 4: Use the formula for the distance from a point to a line in 3D
The distance \(d\) from point \(A\) to the line can be calculated using the formula:
\[ d = \frac{|AB \cdot (D \times AB)|}{|D|} \]
Where:
- \(AB \cdot (D \times AB)\) is the scalar triple product.
- \(D \times AB\) is the cross product of the direction vector and the vector from \(A\) to \(B\).

### Step 5: Calculate \(D \times AB\)
Calculating the cross product:
\[ D \times AB = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 6 \\ -1 & 0 & -3 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}(5 \cdot (-3) - 6 \cdot 0) - \hat{j}(3 \cdot (-3) - 6 \cdot (-1)) + \hat{k}(3 \cdot 0 - 5 \cdot (-1)) \]
\[ = \hat{i}(-15) - \hat{j}(-9 + 6) + \hat{k}(0 + 5) \]
\[ = -15\hat{i} + 3\hat{j} + 5\hat{k} \]
Thus, \(D \times AB = (-15, 3, 5)\).

### Step 6: Calculate the magnitude of \(D\)
\[ |D| = \sqrt{3^2 + 5^2 + 6^2} = \sqrt{9 + 25 + 36} = \sqrt{70} \]

### Step 7: Calculate the scalar triple product \(AB \cdot (D \times AB)\)
\[ AB \cdot (D \times AB) = (-1, 0, -3) \cdot (-15, 3, 5) = (-1)(-15) + (0)(3) + (-3)(5) = 15 + 0 - 15 = 0 \]

### Step 8: Calculate the distance
Since the scalar triple product is zero, the distance \(d\) is:
\[ d = \frac{|0|}{\sqrt{70}} = 0 \]

### Conclusion
The distance of the point \((-2, 4, -5)\) from the line is \(0\), indicating that the point lies on the line.