Given the line with parametric equations $x=1+2t$, $y=3-t$, $z=4+5t$, find the perpendicular distance from the point $(2,1,0)$ to the line.

• A. $\frac{\sqrt{30}}{6}$
• B. $\frac{14}{\sqrt{30}}$
• C. $\frac{\sqrt{30}}{7}$
• D. $\frac{6}{\sqrt{30}}$

Answer: (A)

Verified step by step guidance

1. Identify the direction vector of the line from the parametric equations. Since \(x = 1 + 2t\), \(y = 3 - t\), and \(z = 4 + 5t\), the direction vector is \(\vec{d} = \langle 2, -1, 5 \rangle\).
2. Find a point on the line by substituting \(t = 0\) into the parametric equations, giving the point \(P_0 = (1, 3, 4)\).
3. Form the vector from the point on the line \(P_0\) to the given point \(P = (2, 1, 0)\): \(\vec{v} = \langle 2 - 1, 1 - 3, 0 - 4 \rangle = \langle 1, -2, -4 \rangle\).
4. Calculate the cross product of \(\vec{v}\) and the direction vector \(\vec{d}\): \(\vec{v} \times \vec{d}\), which gives a vector perpendicular to both.
5. Use the formula for the distance from a point to a line in 3D: \(\text{distance} = \frac{\| \vec{v} \times \vec{d} \|}{\| \vec{d} \|}\). Compute the magnitudes of the cross product and the direction vector, then form the ratio to find the perpendicular distance.