Let's solve the problem step-by-step: **Step 1: Definition of Collinear Vectors** Two vectors are collinear if one is a scalar multiple of the other. Let’s check if either \(\vec{b} = [-9, 6]\) or \(\vec{c} = [-6, -4]\) is a scalar multiple of \(\vec{a} = [6, -4]\). For \(\vec{b}\): Suppose \(\vec{b} = k\vec{a}\). \[ [-9, 6] = k[6, -4] \] So, \[ -9 = 6k \implies k = -\frac{3}{2} \] \[ 6 = -4k \implies k = -\frac{3}{2} \] Both components give the same \(k\), so **\(\vec{b}\) is collinear with \(\vec{a}\)**. **Explanation:** \(\vec{b}\) is a scalar multiple of \(\vec{a}\), so they are collinear. --- **Step 2: Check the Second Vector for Collinearity** Now check for \(\vec{c}\): Suppose \(\vec{c} = k\vec{a}\). \[ [-6, -4] = k[6, -4] \] So, \[ -6 = 6k \implies k = -1 \] \[ -4 = -4k \implies k = 1 \] \(k\) is not the same for both components, so **\(\vec{c}\) is not collinear with \(\vec{a}\)**. **Explanation:** \(\vec{c}\) is **not** a scalar multiple of \(\vec{a}\), so they are not collinear. --- ### **Final Answer:** **The vector \(\vec{b} = [-9, 6]\) is collinear with \(\vec{a} = [6, -4]\), because it is a scalar multiple (\(k = -\frac{3}{2}\)) of \(\vec{a}\). Proof: Both components give the same value of \(k\).**