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The line l1 has vector equation \[ r = \begin{pmatrix} -6 \\ 4 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} -4 \\ 3 \\ 1 \end{pmatrix} \] and the line l2 has vector equation \[ r = \begin{pmatrix} -6 \\ 4 \\ 3 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -4 \\ 1 \end{pmatrix} \] where \( \lambda \) and \( \mu \) are parameters - Edexcel - A-Level Maths: Pure - Question 6 - 2010 - Paper 7
Question 6
![The-line-l1-has-vector-equation--\[-r-=-\begin{pmatrix}--6-\\--4-\\---1-\end{pmatrix}-+-\lambda-\begin{pmatrix}--4-\\--3-\\--1-\end{pmatrix}-\]--and-the-line-l2-has-vector-equation--\[-r-=-\begin{pmatrix}--6-\\--4-\\--3-\end{pmatrix}-+-\mu-\begin{pmatrix}-3-\\---4-\\--1-\end{pmatrix}-\]--where-\(-\lambda-\)-and-\(-\mu-\)-are-parameters-Edexcel-A-Level Maths: Pure-Question 6-2010-Paper 7.png](https://cdn.simplestudy.io/assets/backend/uploads/question/MathsPure_2010_7_1_QP_4_2010 .jpg)
The line l1 has vector equation \[ r = \begin{pmatrix} -6 \\ 4 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} -4 \\ 3 \\ 1 \end{pmatrix} \] and the line l2 has ... show full transcript
Worked Solution & Example Answer:The line l1 has vector equation \[ r = \begin{pmatrix} -6 \\ 4 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} -4 \\ 3 \\ 1 \end{pmatrix} \] and the line l2 has vector equation \[ r = \begin{pmatrix} -6 \\ 4 \\ 3 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -4 \\ 1 \end{pmatrix} \] where \( \lambda \) and \( \mu \) are parameters - Edexcel - A-Level Maths: Pure - Question 6 - 2010 - Paper 7
Step 1
a) Write down the coordinates of A.
Answer
To find the coordinates of A, we need to determine the values of ( \lambda ) and ( \mu ) such that the equations for lines l1 and l2 are equal. This gives us the following equations:
-6 + \lambda (-4) = -6 + \mu (3)
4 + \lambda (3) = 4 - \mu (4)
-1 + \lambda (1) = 3 + \mu (1)
Solving these equations simultaneously, we determine the coordinates of A to be ( A(-6, 4, -1) ).
Step 2
b) Find the value of cos θ.
Answer
The direction vectors of the lines are:
For l1: ( \begin{pmatrix} -4 \
3 \
1 \end{pmatrix} ) and for l2: ( \begin{pmatrix} 3 \
-4 \
1 \end{pmatrix} ).
Using the dot product to find ( \cos \theta ):
[ \text{cos} \theta = \frac{\overrightarrow{d_1} \cdot \overrightarrow{d_2}}{|\overrightarrow{d_1}| |\overrightarrow{d_2}|} = \frac{12 + 4 + 3}{\sqrt{(-4)^2 + 3^2 + 1^2} \sqrt{3^2 + (-4)^2 + 1^2}}
= \frac{19}{26}. ]
Step 3
Step 4
Step 5
Step 6
f) Find the length of AY, giving your answer to 3 significant figures.
Answer
Given that ( \overrightarrow{XY} ) is perpendicular to l1, we can use the Pythagorean theorem. The distance d can be calculated by finding the right triangle formed by AY:
Using the previously found values, we can find AY using
[ \frac{4\sqrt{26}}{d} = \cos \theta ]
Thus, the length of AY is:
[ AY = \frac{4}{\sqrt{26}} \times d = 27.9. ]