Mixed practice: lines in 3D, parametric form and intersections Problem set B Answers

1. (a) \(\mathbf{r} = \begin{pmatrix} -1 \\ 4 \\ 6 \end{pmatrix} + \lambda\begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}\)
2. (b) No

1. (a) \(\mathbf{r} = \begin{pmatrix} 0 \\ 5 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} -4 \\ 3 \\ 1 \end{pmatrix}\)
2. (b) \(\mathbf{r} = (6\mathbf{i} + \mathbf{j} - 3\mathbf{k}) + \lambda(\mathbf{i} + 5\mathbf{j} - 2\mathbf{k})\)
3. (c) \(\mathbf{r} = \begin{pmatrix} -4 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 6 \\ -1 \end{pmatrix}\)

1. (a) \(x = -3 + 4\lambda, y = 1 - 6\lambda, z = 5 + 3\lambda\)
2. (b) \(x = 3\lambda, y = 4, z = -2 - \lambda\)
3. (c) \(\mathbf{r} = \begin{pmatrix} -1 \\ 7 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 0 \\ \frac{1}{3} \end{pmatrix}\)

\(x = -2 - \lambda, y = 6 + 4\lambda, z = 3\lambda\)

1. (a) No
2. (b) \(p = -4, q = 13\)

\((-4, 1, 6)\)

1. (a) The third equation fails, so the lines do not intersect.
2. (b) \((1, 5, -3)\)

1. (a) Same direction, and the point is on \(l\) (\(\lambda = -1\)).
2. (b) The direction is parallel but the point is not on \(l\).

\((-3, 5, 2)\)

1. (a) \(\mathbf{r} = \begin{pmatrix} -3 \\ 6 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -3 \\ -4 \end{pmatrix}\)
2. (b) \((1, 0, -6)\) or \((-7, 12, 10)\)

1. (a) \((4, 0, 0)\)
2. (b) The two coordinates give different values of \(\lambda\), so the line does not meet the axis.

\(p = -2\), point \((-2, 4, 3)\)

1. (a) \(13\)
2. (b) \((5, -9, 37)\) or \((-13, 15, -35)\)