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The line l_1 has equation \[ r = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 8
Question 7
The line l_1 has equation \[ r = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} . \] The line l_2 has equation \[ r... show full transcript
Worked Solution & Example Answer:The line l_1 has equation \[ r = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 8
Step 1
Show that l_1 and l_2 do not meet.
Answer
To determine if lines l_1 and l_2 intersect, we can equate their equations:
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From l_1: [ r_1 = \begin{pmatrix} 1 + \lambda \ 0 \ -1 \end{pmatrix} ]
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From l_2: [ r_2 = \begin{pmatrix} 3 + \mu \ 6 - \mu \ 2 - \mu \end{pmatrix} ]
Setting these equal gives us:
[ \begin{pmatrix} 1 + \lambda \ 0 \ -1 \end{pmatrix} = \begin{pmatrix} 3 + \mu \ 6 - \mu \ 2 - \mu \end{pmatrix} ]
This yields three equations:
- ( 1 + \lambda = 3 + \mu )
- ( 0 = 6 - \mu )
- ( -1 = 2 - \mu )
From the second equation, we find ( \mu = 6 ). Substituting ( \mu = 6 ) into the first equation gives:
[ \lambda = 3 + 6 - 1 = 8. ]
The third equation leads to:
[ -1 = 2 - 6 \Longrightarrow -1 = -4 ] (which is incorrect).
This demonstrates that those equations are inconsistent. Thus, l_1 and l_2 do not intersect.
Step 2
Find the cosine of the acute angle between AB and l_1.
Answer
The coordinates of point A on l_1 when ( \lambda = 1 ) are: [ A = \begin{pmatrix} 1 + 1 \ 0 \ -1 \end{pmatrix} = \begin{pmatrix} 2 \ 0 \ -1 \end{pmatrix}. ]
The coordinates of point B on l_2 when ( \mu = 2 ) are: [ B = \begin{pmatrix} 3 + 2 \ 6 - 2 \ 2 - 2 \end{pmatrix} = \begin{pmatrix} 5 \ 4 \ 0 \end{pmatrix}. ]
Next, we calculate the vector ( AB ): [ AB = B - A = \begin{pmatrix} 5 \ 4 \ 0 \end{pmatrix} - \begin{pmatrix} 2 \ 0 \ -1 \end{pmatrix} = \begin{pmatrix} 3 \ 4 \ 1 \end{pmatrix}. ]
The direction vector for l_1 is: [ d_{l_1} = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}. ]
Using the dot product formula: [ \cos \theta = \frac{ AB \cdot d_{l_1}}{|AB| \cdot |d_{l_1}|} ]
Calculating the dot product: [ AB \cdot d_{l_1} = 3 \cdot 1 + 4 \cdot 0 + 1 \cdot 0 = 3. ]
The magnitudes are: [ |AB| = \sqrt{3^2 + 4^2 + 1^2} = \sqrt{26}, \quad |d_{l_1}| = 1. ]
Therefore: [ \cos \theta = \frac{3}{\sqrt{26}}. ]
Thus, the cosine of the acute angle between AB and l_1 is ( \frac{3}{\sqrt{26}} ).