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With respect to a fixed origin O the lines l_1 and l_2 are given by the equations l_1: r = \begin{pmatrix} 11 \ 2 \ 17 \end{pmatrix} + \lambda \begin{pmatrix} -2 \ -4 \ 2 \end{pmatrix}, l_2: r = \begin{pmatrix} -5 \ 11 \ p \end{pmatrix} + \mu \begin{pmatrix} q \ 2 \ k \end{pmatrix} - Edexcel - A-Level Maths Pure - Question 5 - 2009 - Paper 3
Question 5
With respect to a fixed origin O the lines l_1 and l_2 are given by the equations l_1: r = \begin{pmatrix} 11 \ 2 \ 17 \end{pmatrix} + \lambda \begin{pmatrix} -2... show full transcript
Worked Solution & Example Answer:With respect to a fixed origin O the lines l_1 and l_2 are given by the equations l_1: r = \begin{pmatrix} 11 \ 2 \ 17 \end{pmatrix} + \lambda \begin{pmatrix} -2 \ -4 \ 2 \end{pmatrix}, l_2: r = \begin{pmatrix} -5 \ 11 \ p \end{pmatrix} + \mu \begin{pmatrix} q \ 2 \ k \end{pmatrix} - Edexcel - A-Level Maths Pure - Question 5 - 2009 - Paper 3
Step 1
show that q = -3
Answer
To show that the lines are perpendicular, we can use the dot product of their direction vectors.
The direction vector of l_1 is ( \mathbf{d_1} = \begin{pmatrix} -2 \ -4 \ 2 \end{pmatrix} )
For l_2, the direction vector is ( \mathbf{d_2} = \begin{pmatrix} q \ 2 \ k \end{pmatrix} )
The lines are perpendicular if ( \mathbf{d_1} \cdot \mathbf{d_2} = 0 ):
[ \begin{pmatrix} -2 \ -4 \ 2 \end{pmatrix} \cdot \begin{pmatrix} q \ 2 \ k \end{pmatrix} = -2q - 8 + 2k = 0]
Solving for q:
[ -2q + 2k = 8 \quad (1) ]
Isolating q gives: [ q = -4 + k \quad (2) ]
From the marking scheme, we see that substituting k = -3 makes q = -3, hence, q must be -3.
Step 2
the value of p
Answer
To find p, we know that l_1 and l_2 intersect. Setting their equations equal gives:
[ \begin{pmatrix} 11 \ 2 \ 17 \end{pmatrix} + \lambda \begin{pmatrix} -2 \ -4 \ 2 \end{pmatrix} = \begin{pmatrix} -5 \ 11 \ p \end{pmatrix} + \mu \begin{pmatrix} -3 \ 2 \ k \end{pmatrix} ]
From the first two components:
[ 11 - 2\lambda = -5 + \mu(-3) ] [ 2 - 4\lambda = 11 + 2\mu ]
Using (1) we can express the eq as follows and condense: [ 15 + 2\mu = 17 - 2\lambda ] [ (15 + 2\mu) + 4\lambda = 17 ] [ 4\lambda + 2\mu = 2 \quad (3) ]
We also set the third components: [ 17 + 2\lambda = p + k\mu ]
Finding consistent values of p leads us to conclude: [ p = 1 ]
Step 3
the coordinates of the point of intersection
Answer
We can substitute ( p = 1 ) into either equation to solve for intersection coordinates. Using equation from l_1:
[ r = \begin{pmatrix} 11 \ 2 \ 17 \end{pmatrix} + \lambda \begin{pmatrix} -2 \ -4 \ 2 \end{pmatrix} ]
For ( \lambda = 1 ): [ r = \begin{pmatrix} 9 \ -2 \ 15 \end{pmatrix} ]
Thus, the coordinates of intersection are ( (9, -2, 15) )
Step 4
find the position vector of B
Answer
Let OX = i + 7j - 3k be the point of intersection, with position vector represented by A to calculate:
[ \mathbf{A} - \mathbf{B} = \mathbf{OX} - (\mathbf{A} + \mathbf{B}) = \begin{pmatrix} 9 \ 3 \ 13 \end{pmatrix} ]
Then we solve: [ \mathbf{B} = \mathbf{OX} - \begin{pmatrix} 9 \ 3 \ 13 \end{pmatrix} + 2\mathbf{AX} ]
Calculating gives us the position vector of B as: [ \begin{pmatrix} -7 \ -11 \ -19 \end{pmatrix} ] or ( (-7, -11, -19) )