1. **Problem statement:** We are given a parametric vector equation $ \vec{x} = \begin{pmatrix}420 \\ -630 \\ 120 \end{pmatrix} + r \cdot \begin{pmatrix}40 \\ 50 \\ 11 \end{pmatrix} $ and a point $ P(1380, 570, 0) $. The task is to find the parameter $ r $ such that $ \vec{x} = P $. 2. **Formula and approach:** The vector equation represents a line. To find $ r $, set each component equal to the corresponding coordinate of $ P $: $$ \begin{cases} 420 + 40r = 1380 \\ -630 + 50r = 570 \\ 120 + 11r = 0 \end{cases} $$ 3. **Solve each equation for $ r $:** - From the first component: $$ 420 + 40r = 1380 \\ 40r = 1380 - 420 \\ 40r = 960 \\ r = \frac{960}{40} = 24 $$ - From the second component: $$ -630 + 50r = 570 \\ 50r = 570 + 630 \\ 50r = 1200 \\ r = \frac{1200}{50} = 24 $$ - From the third component: $$ 120 + 11r = 0 \\ 11r = -120 \\ r = \frac{-120}{11} \approx -10.91 $$ 4. **Interpretation:** The first two components give $ r = 24 $, but the third gives a different value $ r \approx -10.91 $. This means the point $ P $ does not lie on the line defined by the vector equation. 5. **Conclusion:** There is no single $ r $ that satisfies all three equations simultaneously, so $ P(1380, 570, 0) $ is not on the line. **Final answer:** The point $ P $ is not on the line defined by the vector equation because the parameter $ r $ values do not match for all components.