To find the equation of the perpendicular from the point \( P(2,4,-1) \) to the given line, we need to understand the direction ratios of both the line and the perpendicular.
- The given line is represented as: \(\frac{x+5}{1} = \frac{y+3}{4} = \frac{z-6}{-9}\).
- The direction ratios (dr) of the line are \( 1, 4, -9 \).
- The direction ratios of the perpendicular line, say \( a, b, c \), are such that the dot product with the direction ratios of the given line is zero, as they are perpendicular to each other. \(\Rightarrow a \cdot 1 + b \cdot 4 + c \cdot (-9) = 0\).
- Substitute the point \( P(2,4,-1) \) into the standard form of the line’s equation. A line with direction ratios \( a, b, c \) passing through the point \( (x_1,y_1,z_1) \) is represented as: \(\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}\).
- Here, it is determined that:
- \(x_1 = 2\)
- \(y_1 = 4\)
- \(z_1 = -1\)
- As the perpendicular line is expressed: \(\frac{x-2}{a} = \frac{y-4}{b} = \frac{z+1}{c}\).
- The question provides multiple choices for the direction ratios (which must satisfy the perpendicularity condition in (2)). By verification, the correct direction ratios are \( a = 6 \), \( b = 3 \), and \( c = 2 \), calculated using common sense or trial-error for the values:
- Since \(
a + 4b - 9c = 0\) must be true, a common solution is \( a = 6, b = 3, c = 2 \). Check that: \(6 \cdot 1 + 3 \cdot 4 + 2 \cdot (-9) = 0\), proving \( 6 + 12 - 18 = 0\), confirming perpendicularity.
Thus, the equation of the perpendicular line from \( P(2,4,-1) \) to the given line is:
$\frac{x-2}{6}=\frac{y-4}{3}=\frac{z+1}{2}$
The correct answer is therefore:
$\frac{x-2}{6}=\frac{y-4}{3}=\frac{z+1}{2}$