CAIE
FP1
2002
November
Q9
12 marks
Challenging +1.2
9 The planes \(\Pi _ { 1 }\) and \(\Pi _ { 2 }\), which meet in the line \(/\), have vector equations
$$\begin{aligned}
& \mathbf { r } = 2 \mathbf { i } + 4 \mathbf { j } + 6 \mathbf { k } + \theta _ { 1 } ( 2 \mathbf { i } + 3 \mathbf { k } ) + \phi _ { 1 } ( - 4 \mathbf { j } + 5 \mathbf { k } ) , \\
& \mathbf { r } = 2 \mathbf { i } + 4 \mathbf { j } + 6 \mathbf { k } + \theta _ { 2 } ( 3 \mathbf { j } + \mathbf { k } ) + \phi _ { 2 } ( - \mathbf { i } + \mathbf { j } + 2 \mathbf { k } ) ,
\end{aligned}$$
respectively. Find a vector equation of the line \(l\) in the form \(\mathbf { r } = \mathbf { a } + t \mathbf { b }\).
Find a vector equation of the plane \(\Pi _ { 3 }\) which contains \(l\) and which passes through the point with position vector \(4 \mathbf { i } + 3 \mathbf { j } + 2 \mathbf { k }\). Find also the equation of \(\Pi _ { 3 }\) in the form \(a x + b y + c z = d\).
Deduce, or prove otherwise, that the system of equations
$$\begin{aligned}
& 6 x - 5 y - 4 z = - 32 \\
& 5 x - y + 3 z = 24 \\
& 9 x - 2 y + 5 z = 40
\end{aligned}$$
has an infinite number of solutions.
CAIE
FP1
2013
November
Q6
9 marks
Challenging +1.2
6 The linear transformation \(\mathrm { T } : \mathbb { R } ^ { 4 } \rightarrow \mathbb { R } ^ { 4 }\) is represented by the matrix \(\mathbf { M }\), where
$$\mathbf { M } = \left( \begin{array} { r r r r }
1 & - 3 & - 1 & 2 \\
4 & - 10 & 0 & 2 \\
1 & - 1 & 3 & - 4 \\
5 & - 12 & 1 & 1
\end{array} \right)$$
Find, in either order, the rank of \(\mathbf { M }\) and a basis for the null space \(K\) of T .
Evaluate
$$\mathbf { M } \left( \begin{array} { r }
1 \\
- 2 \\
- 3 \\
- 4
\end{array} \right)$$
and hence show that every solution of
$$\mathbf { M x } = \left( \begin{array} { r }
2 \\
16 \\
10 \\
22
\end{array} \right)$$
has the form
$$\mathbf { x } = \left( \begin{array} { r }
1 \\
- 2 \\
- 3 \\
- 4
\end{array} \right) + \lambda \mathbf { e } _ { 1 } + \mu \mathbf { e } _ { 2 } ,$$
where \(\lambda\) and \(\mu\) are real numbers and \(\left\{ \mathbf { e } _ { 1 } , \mathbf { e } _ { 2 } \right\}\) is a basis for \(K\).