Let’s solve each part step by step using formulas. Given: \(\vec{v} = \langle 7, -2 \rangle\) \(\vec{w} = \langle -1, 4 \rangle\) --- ### a. **Magnitude of** \(\vec{v}\) **Formula:** \[ |\vec{v}| = \sqrt{v_x^2 + v_y^2} \] **Solution:** \[ |\vec{v}| = \sqrt{7^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53} \] --- ### b. **Direction of** \(\vec{v}\) (Direction angle \(\theta\) with respect to the positive x-axis) **Formula:** \[ \theta = \arctan\left(\frac{v_y}{v_x}\right) \] **Solution:** \[ \theta = \arctan\left(\frac{-2}{7}\right) \] --- ### c. **\(\vec{w} - \vec{v}\)** **Formula:** \[ \vec{w} - \vec{v} = \langle w_x - v_x, w_y - v_y \rangle \] **Solution:** \[ \vec{w} - \vec{v} = \langle -1 - 7, \ 4 - (-2) \rangle = \langle -8, 6 \rangle \] --- ### d. **\(\|\vec{w} + \vec{v}\|\)** **Find \(\vec{w} + \vec{v}\) first:** \[ \vec{w} + \vec{v} = \langle -1 + 7, 4 + (-2) \rangle = \langle 6, 2 \rangle \] **Now the magnitude:** \[ \|\vec{w} + \vec{v}\| = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \] --- ### e. **\(\vec{w} \cdot \vec{v}\)** (Dot product) **Formula:** \[ \vec{w} \cdot \vec{v} = w_x v_x + w_y v_y \] **Solution:** \[ \vec{w} \cdot \vec{v} = (-1)(7) + (4)(-2) = -7 - 8 = -15 \] --- ### f. **Angle between** \(\vec{v}\) **and** \(\vec{w}\) **Formula:** \[ \cos\theta = \frac{\vec{v} \cdot \vec{w}}{|\vec{v}| |\vec{w}|} \] First, calculate \(|\vec{w}|\): \[ |\vec{w}| = \sqrt{(-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \] Now plug in values: \[ \cos\theta = \frac{-15}{\sqrt{53}\sqrt{17}} \] \[ \theta = \arccos\left(\frac{-15}{\sqrt{53}\sqrt{17}}\right) \] --- ### g. **\(\vec{k}\) such that it is parallel to \(\vec{w}\)** Any vector parallel to \(\vec{w}\) can be written as \(k\vec{w}\) for some scalar \(k\). \[ \vec{k} = k\langle -1, 4 \rangle \] --- ### h. **\(\vec{m}\) such that it is perpendicular to \(\vec{v}\)** A vector perpendicular to \(\vec{v} = \langle 7, -2 \rangle\) can be written as \(\langle 2, 7 \rangle\) or any scalar multiple. Thus, \[ \vec{m} = m\langle 2, 7 \rangle \] where \(m\) is any real number. --- Let me know if you want numerical values for any part!