9709 P1 - Jun 2007 - Q9
2218
Relative to an origin O , the position vectors of the points A and B are given by
\(\overrightarrow{OA} = \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix}\) and \(\overrightarrow{OB} = \begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix}\).
(i) Given that C is the point such that \(\overrightarrow{AC} = 2\overrightarrow{AB}\), find the unit vector in the direction of \(\overrightarrow{OC}\).
The position vector of the point D is given by \(\overrightarrow{OD} = \begin{pmatrix} 1 \\ 4 \\ k \end{pmatrix}\), where k is a constant, and it is given that \(\overrightarrow{OD} = m\overrightarrow{OA} + n\overrightarrow{OB}\), where m and n are constants.
(ii) Find the values of m , n and k .
Solution
Mark Scheme
Solution
(i) First, find \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ -2 \end{pmatrix}\).
Then, \(\overrightarrow{AC} = 2\overrightarrow{AB} = 2 \begin{pmatrix} -1 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} -2 \\ 2 \\ -4 \end{pmatrix}\).
Now, \(\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} + \begin{pmatrix} -2 \\ 2 \\ -4 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix}\).
The unit vector in the direction of \(\overrightarrow{OC}\) is \(\frac{1}{7} \begin{pmatrix} 2 \\ 3 \\ -6 \end{pmatrix}\).
(ii) We have \(\overrightarrow{OD} = m\overrightarrow{OA} + n\overrightarrow{OB}\).
\(m \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} + n \begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ k \end{pmatrix}\).
This gives the equations:
\(4m + 3n = 1\)
\(m + 2n = 4\)
Solving these simultaneously, we find \(m = -2\) and \(n = 3\).
Substitute \(m\) and \(n\) into the third component equation:
\(-2(-2) + 3(-4) = k\)
\(k = -8\).
