1 Find a vector parallel to the line defined by the parametric equations \( \left\{\begin{array}{l}x(t)=1-9 t \\ y(t)=-1-5 t \\ z(t)=2+6 t\end{array}\right. \) Additionally, find a point on the line. Parallel vector (in angle bracket notation):

To find a vector parallel to the line defined by the parametric equations, we can extract the coefficients of \( t \) from each equation. Here, the parametric equations are: \[ \begin{align*} x(t) &= 1 - 9t, \\ y(t) &= -1 - 5t, \\ z(t) &= 2 + 6t. \end{align*} \] The vector parallel to the line is given by the coefficients of \( t \): \[ \langle -9, -5, 6 \rangle. \] To find a point on the line, we can substitute \( t = 0 \): \[ \begin{align*} x(0) &= 1, \\ y(0) &= -1, \\ z(0) &= 2. \end{align*} \] Thus, a point on the line is \( (1, -1, 2) \). So, the parallel vector is \( \langle -9, -5, 6 \rangle \) and a point on the line is \( (1, -1, 2) \).