Find two vectors parallel to \( \mathbf{v} \) of the given length. \[ v=\langle 9,-12,0\rangle \text {; length }=20 \] The vector in the direction of \( \mathbf{v} \) is \( \langle\square, \square, \square \).

To find a vector parallel to \( \mathbf{v} = \langle 9, -12, 0 \rangle \) with length 20, we first need to determine the unit vector in the direction of \( \mathbf{v} \). The length (or magnitude) of \( \mathbf{v} \) is calculated as: \[ \text{length}(\mathbf{v}) = \sqrt{9^2 + (-12)^2 + 0^2} = \sqrt{81 + 144} = \sqrt{225} = 15. \] Now we find the unit vector \( \mathbf{u} \) by dividing \( \mathbf{v} \) by its length: \[ \mathbf{u} = \frac{\mathbf{v}}{\text{length}(\mathbf{v})} = \left\langle \frac{9}{15}, \frac{-12}{15}, 0 \right\rangle = \left\langle \frac{3}{5}, -\frac{4}{5}, 0 \right\rangle. \] To find vectors parallel to \( \mathbf{v} \) with length 20, we scale the unit vector \( \mathbf{u} \) by 20: \[ \mathbf{parallel1} = 20 \cdot \mathbf{u} = 20 \cdot \left\langle \frac{3}{5}, -\frac{4}{5}, 0 \right\rangle = \left\langle 12, -16, 0 \right\rangle. \] To find another vector in the opposite direction, we can simply negate \( \mathbf{parallel1} \): \[ \mathbf{parallel2} = -20 \cdot \mathbf{u} = -20 \cdot \left\langle \frac{3}{5}, -\frac{4}{5}, 0 \right\rangle = \left\langle -12, 16, 0 \right\rangle. \] Thus, the vectors parallel to \( \mathbf{v} \) of length 20 are: \[ \langle 12, -16, 0 \rangle \quad \text{and} \quad \langle -12, 16, 0 \rangle. \]