Find the (perpendicular) distance from the line given by the parametric equations \[ \begin{array}{l}x(t)=-3-2 t \\ y(t)=3+4 t \\ z(t)=-1-10 t\end{array} \] to the point \( (8,6,-2) \)

To find the perpendicular distance from the given line to the point \( (8, 6, -2) \), we start by expressing the line in vector form. The direction vector of the line can be derived from the coefficients of \( t \) in the parametric equations: \[ \mathbf{d} = \begin{pmatrix} -2 \\ 4 \\ -10 \end{pmatrix} \] The point on the line at parameter \( t \) is: \[ \mathbf{R}(t) = \begin{pmatrix} -3 \\ 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} -2 \\ 4 \\ -10 \end{pmatrix} = \begin{pmatrix} -3 - 2t \\ 3 + 4t \\ -1 - 10t \end{pmatrix} \] We can express the point \( \mathbf{P} = \begin{pmatrix} 8 \\ 6 \\ -2 \end{pmatrix} \) and find the vector from any point on the line to the point \( P \): \[ \mathbf{PQ}(t) = \mathbf{P} - \mathbf{R}(t) = \begin{pmatrix} 8 \\ 6 \\ -2 \end{pmatrix} - \begin{pmatrix} -3 - 2t \\ 3 + 4t \\ -1 - 10t \end{pmatrix} = \begin{pmatrix} 11 + 2t \\ 3 - 4t \\ 9 - 10t \end{pmatrix} \] We want this vector to be perpendicular to the direction vector \( \mathbf{d} \), which means their dot product is zero: \[ \mathbf{PQ}(t) \cdot \mathbf{d} = (11 + 2t)(-2) + (3 - 4t)(4) + (9 - 10t)(-10) = 0 \] Expanding this: \[ -22 - 4t + 12 - 16t - 90 + 100t = 0 \] Combine like terms: \[ (100t - 20t) + (-22 + 12 - 90) = 0 \implies 80t - 100 = 0 \] \[ 80t = 100 \implies t = \frac{100}{80} = \frac{5}{4} \] Now substitute \( t = \frac{5}{4} \) back into the expression for \( \mathbf{R}(t) \) to find the point on the line: \[ \mathbf{R}\left(\frac{5}{4}\right) = \begin{pmatrix} -3 - 2 \left(\frac{5}{4}\right) \\ 3 + 4 \left(\frac{5}{4}\right) \\ -1 - 10 \left(\frac{5}{4}\right) \end{pmatrix} = \begin{pmatrix} -3 - \frac{10}{4} \\ 3 + 5 \\ -1 - \frac{50}{4} \end{pmatrix} = \begin{pmatrix} -3 - 2.5 \\ 8 \\ -1 - 12.5 \end{pmatrix} = \begin{pmatrix} -5.5 \\ 8 \\ -13.5 \end{pmatrix} \] Next, we find the distance from \( \mathbf{R}\left(\frac{5}{4}\right) \) to \( \mathbf{P} \): \[ \mathbf{D} = \mathbf{P} - \mathbf{R}\left(\frac{5}{4}\right) = \begin{pmatrix} 8 \\ 6 \\ -2 \end{pmatrix} - \begin{pmatrix} -5.5 \\ 8 \\ -13.5 \end{pmatrix} = \begin{pmatrix} 8 + 5.5 \\ 6 - 8 \\ -2 + 13.5 \end{pmatrix} = \begin{pmatrix} 13.5 \\ -2 \\ 11.5 \end{pmatrix} \] Now we calculate the distance using the formula: \[ \text{Distance} = \sqrt{(13.5)^2 + (-2)^2 + (11.5)^2} \] Calculating the components: \[ (13.5)^2 = 182.25, \quad (-2)^2 = 4, \quad (11.5)^2 =