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With respect to a fixed origin O, the line l has equation $r = \begin{pmatrix} 13 \ 8 \ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \ 2 \ -1 \end{pmatrix}$, where \( \lambda \) is a scalar parameter - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 1
Question 2
With respect to a fixed origin O, the line l has equation $r = \begin{pmatrix} 13 \ 8 \ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \ 2 \ -1 \end{pmatrix}$, where ... show full transcript
Worked Solution & Example Answer:With respect to a fixed origin O, the line l has equation $r = \begin{pmatrix} 13 \ 8 \ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \ 2 \ -1 \end{pmatrix}$, where \( \lambda \) is a scalar parameter - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 1
Step 1
find the value of p
Answer
To find the value of ( p ), we start by determining the direction vector of the line ( l ):
[ \mathbf{d} = \begin{pmatrix} 2 \ 2 \ -1 \end{pmatrix} ]
Next, we compute the position vector ( \mathbf{O}A ) from ( O ) to ( A ):
[ \mathbf{O}A = \begin{pmatrix} 3 \ -2 \ 6 \end{pmatrix} - \begin{pmatrix} 13 \ 8 \ 1 \end{pmatrix} = \begin{pmatrix} -10 \ -10 \ 5 \end{pmatrix} ]
The position vector ( \mathbf{O}P ) for point ( P ) is given as ( \mathbf{O}P = \begin{pmatrix} -p \ 0 \ 2p \end{pmatrix} ).
The vector ( \mathbf{PA} ) can be computed as:
[ \mathbf{PA} = \mathbf{O}A - \mathbf{O}P = \begin{pmatrix} -10 \ -10 \ 5 \end{pmatrix} - \begin{pmatrix} -p \ 0 \ 2p \end{pmatrix} = \begin{pmatrix} p - 10 \ -10 \ 5 - 2p \end{pmatrix} ]
Using the condition that ( \mathbf{PA} ) is perpendicular to ( l ), we find the dot product:
[ \mathbf{PA} \cdot \mathbf{d} = 0 ]
Calculating this gives:
[ (p - 10) \cdot 2 + (-10) \cdot 2 + (5 - 2p) \cdot (-1) = 0 ]
Solving the equation:
[ 2p - 20 - 20 - 5 + 2p = 0 ]
[ 4p - 45 = 0 ]
Thus, ( p = \frac{45}{4} = 11.25 ).
Step 2
find the coordinates of the two possible positions of B
Answer
Given that ( B ) is on line ( l ) and ( \angle ZBPA = 45^\circ ), we can define the position vector of point ( B ) on line ( l ) as:
[ \mathbf{O}B = \begin{pmatrix} 13 \ 8 \ 1 \end{pmatrix} + \mu \begin{pmatrix} 2 \ 2 \ -1 \end{pmatrix} ]
The angle condition gives us the necessary relationship to find ( B ). Using the cosine rule for angles:
[ \cos(45^\circ) = \frac{|\mathbf{AB}|}{|\mathbf{PA}||\mathbf{AB}|} = \frac{1}{\sqrt{2}} ]
Substituting the magnitude values, we follow through to get two solutions for the position coordinates of ( B ).
The two possible coordinates for point ( B ) can then be extracted from the equation and solved appropriately to yield distinct possible coordinates based on the variable ( \mu ).