Find both the vector equation and the parametric equations of the line through \( (0,0,-6) \) in the direction of the vector \( v=\langle-2,4,0\rangle \), where \( t=0 \) corresponds to the given point. The vector equation is \( \langle x, y, z\rangle=\langle 0,0,-6\rangle+t\langle-2,4,0\rangle \). The parametric equations are \( x=\square, y=\square, z=\square \). (Use the answer from the previous step to find this answer.)

To derive the parametric equations from the given vector equation, we can express the vector equation component-wise. The vector equation is: \[ \langle x, y, z \rangle = \langle 0, 0, -6 \rangle + t \langle -2, 4, 0 \rangle \] This expands to three separate equations for \(x\), \(y\), and \(z\): 1. For \(x\): \[ x = 0 + t(-2) = -2t \] 2. For \(y\): \[ y = 0 + t(4) = 4t \] 3. For \(z\): \[ z = -6 + t(0) = -6 \] Thus, the parametric equations are: \[ x = -2t, \quad y = 4t, \quad z = -6 \]