Problem Solution:

Given:

• Position vector of $A$: $\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}$
• Position vector of $C$: $\begin{pmatrix} 8 \\ 1 \\ -9 \end{pmatrix}$

(a) Express $\vec{AC}$ as a column vector:

The vector $\vec{AC}$ is given by: $\vec{AC} = \vec{C} - \vec{A}$ $\vec{AC} = \begin{pmatrix} 8 \\ 1 \\ -9 \end{pmatrix} - \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} = \begin{pmatrix} 8 - 2 \\ 1 - 4 \\ -9 - 6 \end{pmatrix} = \begin{pmatrix} 6 \\ -3 \\ -15 \end{pmatrix}$ Thus, $\vec{AC} = \begin{pmatrix} 6 \\ -3 \\ -15 \end{pmatrix}$.

(b) Find the position vector of $B$:

Point $B$ lies on $AC$, meaning $\vec{B}$ can be expressed as: $\vec{B} = \vec{A} + \lambda \vec{AC}$ where $\lambda$ is a scalar parameter.

Since $AB$ is perpendicular to $OB$, the dot product $\vec{AB} \cdot \vec{OB} = 0$. First, we compute: $\vec{AB} = \vec{B} - \vec{A} = (\vec{A} + \lambda \vec{AC}) - \vec{A} = \lambda \vec{AC}$ and $\vec{OB} = \vec{B} = \vec{A} + \lambda \vec{AC}$

The perpendicularity condition becomes: $\vec{AB} \cdot \vec{OB} = (\lambda \vec{AC}) \cdot (\vec{A} + \lambda \vec{AC}) = 0$

Substitute $\vec{AC} = \begin{pmatrix} 6 \\ -3 \\ -15 \end{pmatrix}$ and $\vec{A} = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}$, and solve for $\lambda$. Expand the dot product: $\lambda \vec{AC} \cdot \vec{A} + \lambda^2 \vec{AC} \cdot \vec{AC} = 0$

1. Compute $\vec{AC} \cdot \vec{A}$: $\vec{AC} \cdot \vec{A} = \begin{pmatrix} 6 \\ -3 \\ -15 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} = (6 \cdot 2) + (-3 \cdot 4) + (-15 \cdot 6) = 12 - 12 - 90 = -90$
2. Compute $\vec{AC} \cdot \vec{AC}$: $\vec{AC} \cdot \vec{AC} = \begin{pmatrix} 6 \\ -3 \\ -15 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ -3 \\ -15 \end{pmatrix} = (6 \cdot 6) + (-3 \cdot -3) + (-15 \cdot -15) = 36 + 9 + 225 = 270$

The equation becomes: $\lambda (-90) + \lambda^2 (270) = 0$ Factorize: $\lambda (-90 + 270\lambda) = 0$ $\lambda = 0 \quad \text{or} \quad \lambda = \frac{1}{3}$

When $\lambda = 0$, $\vec{B} = \vec{A}$, but $B \neq A$. Thus, $\lambda = \frac{1}{3}$.

Substitute $\lambda = \frac{1}{3}$ into $\vec{B} = \vec{A} + \lambda \vec{AC}$: $\vec{B} = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} + \frac{1}{3} \begin{pmatrix} 6 \\ -3 \\ -15 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} + \begin{pmatrix} 2 \\ -1 \\ -5 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix}$ Thus, $\vec{B} = \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix}$.

(c) Find the shortest distance from $O$ to the line $AB$:

The shortest distance from a point to a line is given by: $d = \frac{|\vec{OA} \times \vec{AB}|}{|\vec{AB}|}$

1. Compute $\vec{OA} = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}$, $\vec{AB} = \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} - \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ -5 \end{pmatrix}$.

2. Compute $\vec{OA} \times \vec{AB}$: [ \vec{OA} \times \vec{AB} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 4 & 6 \ 2 &