You're on a roll — let's solve these two problems. Problem 8: Vector Midpoint Step 1: Express the vector $\vec{BC}$ in terms of the given vectors $\vec{AB}$ and $\vec{AC}$. We know that $\vec{BC} = \vec{BA} + \vec{AC}$. Since $\vec{BA} = -\vec{AB}$, we have: $$ \vec{BC} = -\vec{AB} + \vec{AC} $$ Substitute the given vectors: $$ \vec{BC} = -\begin{pmatrix} 6 \\ 4 \end{pmatrix} + \begin{pmatrix} 2 \\ -4 \end{pmatrix} $$ $$ \vec{BC} = \begin{pmatrix} -6 \\ -4 \end{pmatrix} + \begin{pmatrix} 2 \\ -4 \end{pmatrix} $$ $$ \vec{BC} = \begin{pmatrix} -6 + 2 \\ -4 - 4 \end{pmatrix} $$ $$ \vec{BC} = \begin{pmatrix} -4 \\ -8 \end{pmatrix} $$ Step 2: Calculate the position vector of the midpoint M of BC relative to point A. Let M be the midpoint of BC. The position vector of M relative to A, denoted $\vec{AM}$, can be found using the formula: $$ \vec{AM} = \frac{1}{2}(\vec{AB} + \vec{AC}) $$ Substitute the given vectors: $$ \vec{AM} = \frac{1}{2} \left( \begin{pmatrix} 6 \\ 4 \end{pmatrix} + \begin{pmatrix} 2 \\ -4 \end{pmatrix} \right) $$ $$ \vec{AM} = \frac{1}{2} \begin{pmatrix} 6 + 2 \\ 4 - 4 \end{pmatrix} $$ $$ \vec{AM} = \frac{1}{2} \begin{pmatrix} 8 \\ 0 \end{pmatrix} $$ $$ \vec{AM} = \begin{pmatrix} 4 \\ 0 \end{pmatrix} $$ The midpoint of $\vec{BC}$ (represented as the position vector from A) is $\boxed{\begin{pmatrix} 4 \\ 0 \end{pmatrix}}$. Problem 9: Total Surface Area of a Cone Step 1: Identify the given values for the cone. Radius of the base, $r = OC = 7$ cm. Slanting height, $l = AC = 21$ cm. Step 2: Recall the formula for the total surface area of a cone. The total surface area ($TSA$) of a cone is the sum of the area of its circular base and its curved surface area. $$ TSA = \pi r^2 + \pi r l $$ This formula can also be written as: $$ TSA = \pi r (r + l) $$ Step 3: Substitute the given values into the formula and calculate the total surface area. $$ TSA = \pi (7 \text{ cm}) (7 \text{ cm} + 21 \text{ cm}) $$ $$ TSA = \pi (7 \text{ cm}) (28 \text{ cm}) $$ $$ TSA = 196\pi \text{ cm}^2 $$ To get a numerical value, use $\pi \approx 3.14159$: $$ TSA \approx 196 \times 3.14159 $$ $$ TSA \approx 615.75244 \text{ cm}^2 $$ Rounding to two decimal places: $$ TSA \approx 615.75 \text{ cm}^2 $$ The total surface area of the cone is $\boxed{615.75 \text{ cm}^2}$. That's all the problems from the image! What's next?