Photo AI
With respect to a fixed origin O, the line l is given by the equation r = \begin{pmatrix} 8 \\ 1 \\ -3 \end{pmatrix} + \mu \begin{pmatrix} -5 \\ 4 \\ 3 \end{pmatrix} where \mu is a scalar parameter - Edexcel - A-Level Maths Pure - Question 2 - 2015 - Paper 4
Question 2
With respect to a fixed origin O, the line l is given by the equation r = \begin{pmatrix} 8 \\ 1 \\ -3 \end{pmatrix} + \mu \begin{pmatrix} -5 \\ 4 \\ 3 \end{pmatr... show full transcript
Worked Solution & Example Answer:With respect to a fixed origin O, the line l is given by the equation r = \begin{pmatrix} 8 \\ 1 \\ -3 \end{pmatrix} + \mu \begin{pmatrix} -5 \\ 4 \\ 3 \end{pmatrix} where \mu is a scalar parameter - Edexcel - A-Level Maths Pure - Question 2 - 2015 - Paper 4
Step 1
Find the coordinates of A.
Answer
To find the coordinates of point A, we substitute \mu = 1 into the equation for the line l:
r = \begin{pmatrix} 8 \\ 1 \\ -3 \end{pmatrix} + 1 \begin{pmatrix} -5 \\ 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 8 - 5 \\ 1 + 4 \\ -3 + 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 5 \\ 0 \end{pmatrix}$$ Thus, the coordinates of A are (3, 5, 0).Step 2
Write down a vector equation for the line l2.
Answer
The direction vector for line l1 is \begin{pmatrix} -5 \ 4 \ 3 \end{pmatrix}. Since line l2 is parallel to l1 and passes through point P, the vector equation for l2 can be expressed as:
l_2: \mathbf{r} = \begin{pmatrix} 1 \\ 5 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -5 \\ 4 \\ 3 \end{pmatrix}$$ where \lambda is a scalar parameter.Step 3
Find the exact value of the distance AP. Give your answer in the form k√2, where k is a constant to be determined.
Answer
To find the distance AP, we first need to compute the position vector of A and P:
5 \\ 0 \end{pmatrix} - \begin{pmatrix} 1 \\ 5 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ -2 \end{pmatrix} $$ Then, we calculate the magnitude of vector AP: $$ |AP| = \sqrt{(2)^2 + (0)^2 + (-2)^2} = \sqrt{4 + 0 + 4} = \sqrt{8} = 2\sqrt{2} $$ Thus, k = 2.Step 4
Find the value of cosθ.
Answer
To find cosθ, we utilize the dot product between vectors AP and the direction vector of l1:
4 \\ 3 \end{pmatrix} $$ The formula for the cosine of the angle between two vectors is: $$ cos(\theta) = \frac{AP \cdot \mathbf{d}}{|AP| |\mathbf{d}|} $$ Calculating the dot product: $$ AP \cdot \mathbf{d} = \begin{pmatrix} 2 \\ 0 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} -5 \\ 4 \\ 3 \end{pmatrix} = 2\cdot(-5) + 0\cdot4 + (-2)\cdot3 = -10 - 6 = -16 $$ The magnitude of the direction vector: $$ |\mathbf{d}| = \sqrt{(-5)^2 + (4)^2 + (3)^2} = \sqrt{25 + 16 + 9} = \sqrt{50} $$ So, $$ cos(\theta) = \frac{-16}{2\sqrt{2} \times \sqrt{50}} = \frac{-16}{14.142} = -\frac{8}{7\sqrt{2}} $$Step 5
Find the coordinates of the two possible positions of E.
Answer
Given that AP = PE, we can expressE in terms of P. The coordinates of E will be:
4 \\ 3 \end{pmatrix} $$ Thus, substituting in the coordinates of P: $$ E = \begin{pmatrix} 1 \\ 5 \\ 2 \end{pmatrix} + k \begin{pmatrix} -5 \\ 4 \\ 3 \end{pmatrix} $$ This will yield two possible positions depending on the value of k: 1. For k = 1: $$ E_1 = \begin{pmatrix} 1 - 5 \\ 5 + 4 \\ 2 + 3 \end{pmatrix} = \begin{pmatrix} -4 \\ 9 \\ 5 \end{pmatrix} $$ 2. For k = -1: $$ E_2 = \begin{pmatrix} 1 + 5 \\ 5 - 4 \\ 2 - 3 \end{pmatrix} = \begin{pmatrix} 6 \\ 1 \\ -1 \end{pmatrix} $$ Thus, the two possible positions of E are (-4, 9, 5) and (6, 1, -1).