Let's clarify and analyze the problem step by step. First, let's try to rewrite the original line in standard parametric vector form. The original statement is a bit jumbled, but it appears to be: **Given line:** \[ \vec{r}(t) = (5 - 3t)\vec{i} + (6 + t)\vec{j} + (2t)\vec{k} \] Or, written as a point plus a multiple of a direction vector: \[ \vec{r}(t) = \langle 5, 6, 0 \rangle + t \langle -3, 1, 2 \rangle \] --- ### Step 1: Parse Each Option Let's rewrite each option in parametric form and compare the point and direction vector to the given line. #### Option A: \[ \vec{r}_A(t) = \langle 2, 7, 2 \rangle + t\langle -3, 1, 2 \rangle \] - **Explanation:** The direction vector matches the given line (\(-3, 1, 2\)). The point \((2, 7, 2)\) may or may not be on the line. #### Option B: \[ \vec{r}_B(t) = \langle 7 + t, 2 + 2t, 0 \rangle \] - **Explanation:** Direction vector is \(\langle 1, 2, 0 \rangle\) which does **not** match the given direction vector. #### Option C: \[ \vec{r}_C(t) = \langle -5 - t, 9 + 5t, 6 + 10t \rangle \] - **Explanation:** Direction vector is \(\langle -1, 5, 10 \rangle\). Let's check if this is a multiple of \(\langle -3, 1, 2 \rangle\): \[ \frac{-1}{-3} = \frac{1}{3} \\ \frac{5}{1} = 5 \\ \frac{10}{2} = 5 \] No, not a constant multiple. So this does **not** match. #### Option D: \[ \vec{r}_D(t) = \langle -7 + 9t, -7 - 4t, 8 - 6t \rangle \] - **Explanation:** Direction vector is \(\langle 9, -4, -6 \rangle\), which is not a multiple of \(\langle -3, 1, 2 \rangle\). #### Option E: \[ \vec{r}_E(t) = \langle -1 - 9t, 8 + 3t, 4 + 6t \rangle \] - **Explanation:** Direction vector is \(\langle -9, 3, 6 \rangle\). Let's check if this is a constant multiple: \[ \frac{-9}{-3} = 3 \\ \frac{3}{1} = 3 \\ \frac{6}{2} = 3 \] Yes! So this is a multiple of the original direction vector. #### Option F: \[ \vec{r}_F(t) = \langle 10 + 15t, 12 - 5t, 10 - 17t \rangle \] - **Explanation:** Direction vector is \(\langle 15, -5, -17 \rangle\), which is not a constant multiple of \(\langle -3, 1, 2 \rangle\). --- ### Step 2: Check if the Point is on the Line Now for options A and E (the only ones with the correct direction vector or its multiple), check if the given point can be reached by some value of \(t\): #### Option A: Point: \((2, 7, 2)\) Solve for \(t\) in the original line: \[ 5 - 3t = 2 \implies t = 1 \\ 6 + t = 7 \implies t = 1 \\ 0 + 2t = 2 \implies t = 1 \] All are consistent for \(t = 1\). So, Option A **is equivalent**. #### Option E: Point: \((-1, 8, 4)\) Is there a \(t\) so that: \[ 5 - 3t = -1 \implies t = 2 \\ 6 + t = 8 \implies t = 2 \\ 0 + 2t = 4 \implies t = 2 \] All are consistent for \(t = 2\). So, Option E **is equivalent**. --- ## **Final Answer** **The parameterized lines that represent the same line as the given one are:** \[ \boxed{ \text{Option A: } \langle 2, 7, 2 \rangle + t\langle -3, 1, 2 \rangle \\ \text{Option E: } \langle -1, 8, 4 \rangle + t\langle -9, 3, 6 \rangle } \] --- **Summary:** - Step 1: Identified the direction vectors of each option and compared them to the given line's direction. - Step 2: Checked if the point in each valid option lies on the given line by solving for \(t\). - **Final Answer:** Options **A** and **E** represent the same line as the original.