2024 ACJC Promo Q11

The line \(l_1\) has the vector equation \(\mathbf{r} = \begin{pmatrix} a \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 1 \\ b \end{pmatrix}\), \(\lambda \in \mathbb{R}\), where \(a\) and \(b\) are real constants.

The plane \(p_1\) has cartesian equation \(x + 2y + 4z = 4\).

The line \(l_1\) is perpendicular to the line \(l_2\) with cartesian equation \(\frac{1-y}{2} = z - 3\), \(x = 5\).

The line \(l_1\) intersects the plane \(p_1\) at the point with coordinates \((a, 1, 0)\).

1. By first finding the vector equation of line \(l_2\), show that \(a = b = 2\).[3]
2. Verify that the point \(A\) with coordinates \((2, 2, 2)\) lies on \(l_1\).[1]
3. Find the coordinates of the foot of perpendicular, \(F\), from point \(A\) to the plane \(p_1\).
   Hence find the shortest distance from point \(A\) to the plane \(p_1\).[5]
4. The plane \(p_2\) contains the line \(l_1\) and the point \(F\). Find the vector equation of the plane \(p_2\), giving your answer in the form \(\mathbf{r} \cdot \mathbf{n} = d\), where \(\mathbf{n}\) is a vector and \(d\) is a real constant to be determined.[2]