Challenging +1.2 This is a two-part Further Maths vectors question requiring finding a line parallel to two planes (using cross product of direction vectors) and then computing shortest distance between skew lines. While it involves multiple steps and Further Maths content, the techniques are standard and methodical: cross product for direction, then either using the scalar triple product formula or projection methods for distance. The computational work is moderate but straightforward with no novel geometric insight required.
The planes \(\Pi_1\) and \(\Pi_2\) have vector equations
$$\mathbf{r} = \lambda_1(\mathbf{i} + \mathbf{j} - \mathbf{k}) + \mu_1(2\mathbf{i} - \mathbf{j} + \mathbf{k}) \quad \text{and} \quad \mathbf{r} = \lambda_2(\mathbf{i} + 2\mathbf{j} + \mathbf{k}) + \mu_2(3\mathbf{i} + \mathbf{j} - \mathbf{k})$$
respectively. The line \(l\) passes through the point with position vector \(4\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}\) and is parallel to both \(\Pi_1\) and \(\Pi_2\). Find a vector equation for \(l\). [6]
Find also the shortest distance between \(l\) and the line of intersection of \(\Pi_1\) and \(\Pi_2\). [4]
The planes $\Pi_1$ and $\Pi_2$ have vector equations
$$\mathbf{r} = \lambda_1(\mathbf{i} + \mathbf{j} - \mathbf{k}) + \mu_1(2\mathbf{i} - \mathbf{j} + \mathbf{k}) \quad \text{and} \quad \mathbf{r} = \lambda_2(\mathbf{i} + 2\mathbf{j} + \mathbf{k}) + \mu_2(3\mathbf{i} + \mathbf{j} - \mathbf{k})$$
respectively. The line $l$ passes through the point with position vector $4\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}$ and is parallel to both $\Pi_1$ and $\Pi_2$. Find a vector equation for $l$. [6]
Find also the shortest distance between $l$ and the line of intersection of $\Pi_1$ and $\Pi_2$. [4]
\hfill \mbox{\textit{CAIE FP1 2005 Q9 [10]}}