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

1. (a) \(\mathbf{r} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}\)
2. (b) Yes: \(\lambda = -2\)

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

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

\(x = 4 + 3\lambda, y = -1 - \lambda, z = 3 + 2\lambda\)

1. (a) Yes: \(\lambda = -2\)
2. (b) \(p = 11, q = 1\)

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

1. (a) The third equation fails, so the lines do not intersect.
2. (b) \((2, -1, 6)\)

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

\((4, 2, -1)\)

1. (a) \(\mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}\)
2. (b) \((11, 5, -8)\) or \((-7, -7, 16)\)

1. (a) \((0, 5, 0)\)
2. (b) The two coordinates give different values of \(\lambda\), so the line does not meet the axis.

\(p = 4\), point \((5, 1, 2)\)

1. (a) \(11\)
2. (b) \((11, 23, -34)\) or \((-5, -25, 38)\)