The distance of the line \( \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4} \) from the point \( (1, 4, 0) \) along the line \( \frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} \) is:
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When finding the distance between a point and a line in 3D, use the cross product of the vector from the point to any point on the line with the direction vector of the line, and divide by the magnitude of the direction vector of the line.
To determine the distance of the line given by the symmetric equations \( \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4} \) from the point \( (1, 4, 0) \) along the line represented by \( \frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} \), the following steps are performed:
Construct the equation of a plane that encompasses the point \( (1, 4, 0) \) and is perpendicular to the line \( \frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} \). The direction ratios of this line, (1, 2, 3), serve as the normal vector for the plane.
The plane equation, with a normal vector (1, 2, 3) and passing through \( (1, 4, 0) \), is formulated as: \( 1(x - 1) + 2(y - 4) + 3(z - 0) = 0 \).
Upon simplification, the plane equation is derived as \( x + 2y + 3z = 9 \).
Determine the direction ratios of the line \( \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4} \), which are (2, 3, 4). A point on this line is \( (2, 6, 3) \).
Substitute the point \( (2, 6, 3) \) into the plane equation \( x + 2y + 3z = 9 \) to verify its position relative to the plane.
Calculation: \( 2(1) + 6(2) + 3(3) \) = 2 + 12 + 9 = 23, which is not equal to 9.
Since the point does not lie on the plane, calculate the distance from the plane to the line using the formula \( \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \), where \( (x_1, y_1, z_1) \) is the point \( (2, 6, 3) \).
