Find both the vector equation and the parametric equations of the line through \( (4,3,4) \) that is perpendicular to the lines \( x=7-2 t, y=4+8 t, z=4-2 t \) and \( x=-2 t, y=4+t, z=4-t \), where \( t=0 \) corresponds to the given point. The vector equation is \( \langle x, y, z\rangle=\langle \)

To find the vector equation of the line through the point \( (4,3,4) \) that is perpendicular to the two given lines, we first need to identify direction vectors for each line. The direction vector for the first line \( x=7-2t, y=4+8t, z=4-2t \) can be determined from the coefficients of \( t \): \[ \mathbf{d_1} = \langle -2, 8, -2 \rangle \] For the second line \( x=-2t, y=4+t, z=4-t \), the direction vector is: \[ \mathbf{d_2} = \langle -2, 1, -1 \rangle \] Next, we find a vector that is perpendicular to both \( \mathbf{d_1} \) and \( \mathbf{d_2} \) using the cross product: \[ \mathbf{n} = \mathbf{d_1} \times \mathbf{d_2} \] Calculating this gives: \[ \mathbf{n} = \langle -2, 8, -2 \rangle \times \langle -2, 1, -1 \rangle = \langle 6, 2, -14 \rangle \] Now we can write the vector equation of the line through the point \( (4, 3, 4) \) in the direction of \( \mathbf{n} \): \[ \langle x, y, z \rangle = \langle 4, 3, 4 \rangle + t \langle 6, 2, -14 \rangle \] So, the vector equation is: \[ \langle x, y, z\rangle=\langle 4, 3, 4 \rangle + t \langle 6, 2, -14 \rangle \] Now, we can find the parametric equations by expressing \( x, y, z \) in terms of \( t \): \[ x = 4 + 6t, \quad y = 3 + 2t, \quad z = 4 - 14t \] Thus, the parametric equations are: \[ x = 4 + 6t, \quad y = 3 + 2t, \quad z = 4 - 14t \]