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. The lines l1 and l2 intersect at the point A and the acute angle between l1 and l2 is $ \theta $. a) Write down the coordinates of A. b) Find the value of $ \cos \theta $. The point X lies on l1, where $ \lambda = 4 $. c) Find the coordinates of X. d) Find the vector $ \overrightarrow{AX} $. e) Hence, or otherwise, show that $ |\overrightarrow{AX}| = 4\sqrt{26}. $ The point Y lies on l2. Given that the vector $ \overrightarrow{XY} $ is perpendicular to l1, f) Find the length of AY, giving your answer to 3 significant figures.

A-Level Edexcel Maths Pure Differentiation: The line l1 has vector equation

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$

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.

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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.

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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

c) Find the coordinates of X.

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For the point X on l1 where ( \lambda = 4 ):
[ X = \begin{pmatrix} -6 \ 4 \ -1 \end{pmatrix} + 4 \begin{pmatrix} -4 \ 3 \ 1 \end{pmatrix}
= \begin{pmatrix} 10 \ 0 \ 11 \end{pmatrix} ]

Step 4

d) Find the vector AX.

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To find the vector ( \overrightarrow{AX} ):
[ \overrightarrow{AX} = \begin{pmatrix} 10 \ 0 \ 11 \end{pmatrix} - \begin{pmatrix} -6 \ 4 \ -1 \end{pmatrix}
= \begin{pmatrix} 16 \ -4 \ 12 \end{pmatrix} ]

Step 5

e) Hence, or otherwise, show that |AX| = 4√26.

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To find the magnitude of ( \overrightarrow{AX} ):
[ |\overrightarrow{AX}| = \sqrt{16^2 + (-4)^2 + 12^2} = \sqrt{256 + 16 + 144} = \sqrt{416} = 4\sqrt{26}. ]

Step 6

f) Find the length of AY, giving your answer to 3 significant figures.

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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. ]

a) Write down the coordinates of A.

b) Find the value of cos θ.

c) Find the coordinates of X.

d) Find the vector AX.

e) Hence, or otherwise, show that |AX| = 4√26.

f) Find the length of AY, giving your answer to 3 significant figures.

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