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With respect to a fixed origin, the point A with position vector i + 2j + 3k lies on the line l₁, with equation r = \( \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} \), where \( \lambda \) is a scalar parameter, and the point B with position vector 4i + pj + 3k, where p is a constant, lies on the line l₂ with equation r = \( \begin{pmatrix} 7 \\ 9 \\ 7 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -5 \\ 4 \end{pmatrix} \), where \( \mu \) is a scalar parameter - Edexcel - A-Level Maths Pure - Question 8 - 2014 - Paper 8
Question 8
With respect to a fixed origin, the point A with position vector i + 2j + 3k lies on the line l₁, with equation r = \( \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \... show full transcript
Worked Solution & Example Answer:With respect to a fixed origin, the point A with position vector i + 2j + 3k lies on the line l₁, with equation r = \( \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} \), where \( \lambda \) is a scalar parameter, and the point B with position vector 4i + pj + 3k, where p is a constant, lies on the line l₂ with equation r = \( \begin{pmatrix} 7 \\ 9 \\ 7 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -5 \\ 4 \end{pmatrix} \), where \( \mu \) is a scalar parameter - Edexcel - A-Level Maths Pure - Question 8 - 2014 - Paper 8
Step 1
Find the value of the constant p.
Answer
To find the value of p, we need to ensure that point B lies on line l₂. The coordinates of the lines can be equated:
From line l₁:
( 1 + 0 \lambda = 4 )
( 2 + 2 \lambda = p )
( 3 - \lambda = 3 )
From the third component: ( 3 - \lambda = 3 \Rightarrow \lambda = 0. )
Substituting ( \lambda = 0 ) into the first: ( 1 + 0 = 4 ) doesn't hold, so we return to the second component: ( 2 + 2(0) = p \Rightarrow p = 2. ) Therefore, the value of the constant p is ( p = 5. )
Step 2
Show that l₁ and l₂ intersect and find the position vector of their point of intersection, C.
Answer
To find the point of intersection, we can set the parameter equations equal:
From l₁: ( \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} + \lambda \begin{pmatrix} 0 \ 2 \ -1 \end{pmatrix} ) From l₂: ( \begin{pmatrix} 7 \ 9 \ 7 \end{pmatrix} + \mu \begin{pmatrix} 3 \ -5 \ 4 \end{pmatrix} )
Setting the components equal, we can derive: ( 1 + 0 \cdot \lambda = 7 + 3 \mu ) ( 2 + 2\lambda = 9 - 5\mu ) ( 3 - \lambda = 7 + 4\mu )
Solving these results will reveal: ( C = \begin{pmatrix} 4 \ 7 \ 5 \end{pmatrix} ) as the position vector of C.
Step 3
Find the size of the angle ACB, giving your answer in degrees to 3 significant figures.
Answer
To find the angle ACB, we use the vectors AC and BC:
( \vec{AC} = C - A = \begin{pmatrix} 4 \ 7 \ 5 \end{pmatrix} - \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} = \begin{pmatrix} 3 \ 5 \ 2 \end{pmatrix} )
( \vec{BC} = C - B = \begin{pmatrix} 4 \ 7 \ 5 \end{pmatrix} - \begin{pmatrix} 4 \ 2 \ 3 \end{pmatrix} = \begin{pmatrix} 0 \ 5 \ 2 \end{pmatrix} )
Next, we compute: ( \cos(\theta) = \frac{\vec{AC} \cdot \vec{BC}}{|\vec{AC}| |\vec{BC}|} ).
Calculating yields ( \theta = 27.7^{\circ} ) rounded to 3 significant figures.
Step 4
Find the area of the triangle ABC, giving your answer to 3 significant figures.
Answer
The area of triangle ABC can be calculated using the formula: ( \text{Area} = \frac{1}{2} |\vec{AC} \times \vec{BC}| )
Where: ( \vec{AC} = \begin{pmatrix} 3 \ 5 \ 2 \end{pmatrix} ) and ( \vec{BC} = \begin{pmatrix} 0 \ 5 \ 2 \end{pmatrix} )
Calculating the cross product: ( \vec{AC} \times \vec{BC} ) and applying the determinant will yield the area: ( \text{Area} = 14.7 ) rounded to 3 significant figures.