1 problem found
State carefully the conditions which the fixed vectors \(\mathbf{a,b,u}\) and \(\mathbf{v}\) must satisfy in order to ensure that the line \(\mathbf{r=a+}\lambda\mathbf{u}\) intersects the line \(\mathbf{r=b+\mu}\mathbf{v}\) in exactly one point. Find the two values of the fixed scalar \(b\) for which the planes with equations \[ \left.\begin{array}{c} x+y+bz=b+2\\ bx+by+z=2b+1 \end{array}\right\} \tag{*} \] do not intersect in a line. For other values of \(b\), express the line of intersection of the two planes in the form \(\mathbf{r=a}+\lambda\mathbf{u},\) where \(\mathbf{a\cdot u}=0\). Find the conditions which \(b\) and the fixed scalars \(c\) and \(d\) must satisfy to ensure that there is exactly one point on the line \[ \mathbf{r=}\left(\begin{array}{c} 0\\ 0\\ c \end{array}\right)+\mu\left(\begin{array}{c} 1\\ d\\ 0 \end{array}\right) \] whose coordinates satisfy both equations \((*)\).
Solution: There are two requirements (assuming they are lines not fixed points): 1. They cannot be parallel, ie \(\mathbf{u} \neq \lambda \mathbf{v}\) for any \(\lambda\) 2. They must lie in the same plane, ie \((\mathbf{b}-\mathbf{a})\cdot (\mathbf{u} \times \mathbf{v}) = 0\) The planes will not intersect in a line if they are either parallel and separate or parallel and the same. If \(b = 1\) or \(b=-1\) the planes are parallel. \begin{align*} && (x+y) + b z &= b+ 2\\ &&b(x+y) + z &= 2b + 1 \\ \Rightarrow && (1-b^2)z &= 2b+1 - b^2 -2b \\ &&&= 1-b^2 \\ \Rightarrow && z &= 1 \\ && x+ y &= 2 \\ \end{align*} Therefore our line is \(\mathbf{r} = \begin{pmatrix} 1+t \\ 1-t \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \) We must have: \(d \neq -1\) to ensure that the lines aren't parallel. We must also have: \begin{align*} 0 &= \left ( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} -\begin{pmatrix} 0 \\ 0 \\ c \end{pmatrix}\right) \cdot \left ( \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ d \\ 0 \end{pmatrix} \right) \\ &= \begin{pmatrix} 1 \\ 1 \\ 1-c \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ d+1 \end{pmatrix} \\ &= (1-c)(d+1) \end{align*} So \(c =1\)