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. The point A lies on l and has coordinates (3, -2, 6). The point P has position vector (-p i + 2 p k) relative to O, where p is a constant. Given that vector $ \vec{PA} $ is perpendicular to l, a) find the value of p. Given also that B is a point on l such that $ \angle BAP = 45^\circ $, b) find the coordinates of the two possible positions of B.

A-Level Edexcel Maths Pure Integration: With respect to a fixed origin O

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 3 - 2013 - Paper 9

Question 3

$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 3-2013-Paper 9.png$

With respect to a fixed origin O, the line l has equation $$r = \begin{pmatrix} 13 \\ 8 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix},$$ w... 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 3 - 2013 - Paper 9

Step 1

find the value of p.

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Answer

To determine the value of ( p ), we first express the position vectors.

The position vector of point A is: $\vec{OA} = \begin{pmatrix} 3 \\ -2 \\ 6 \end{pmatrix}$

The position vector of point P is: $\vec{OP} = \begin{pmatrix} -p \\ 0 \\ 2p \end{pmatrix}$

The direction vector of line l is: $\vec{d} = \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}$

The vector ( \vec{PA} ) can be found by: $\vec{PA} = \vec{OA} - \vec{OP} = \begin{pmatrix} 3 + p \\ -2 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ -2p \end{pmatrix} = \begin{pmatrix} 3 + p \\ -2 \\ 6 - 2p \end{pmatrix}$

Since ( \vec{PA} ) is perpendicular to ( \vec{d} ), their dot product must equal zero:

$\vec{PA} \cdot \vec{d} = 0$

Calculating the dot product: $\begin{pmatrix} 3 + p \\ -2 \\ 6 - 2p \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix} = 0$ \begin{align*} 2(3 + p) + 2(-2) - (6 - 2p) &= 0
6 + 2p - 4 - 6 + 2p &= 0
4p - 4 &= 0
4p &= 4
p &= 1 \end{align*}

Step 2

find the coordinates of the two possible positions of B.

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Answer

To find coordinates of point B, we note that it lies on line l. The equation for line l can be given with parameter ( \mu ):

$r = \begin{pmatrix} 13 \\ 8 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}$

To express B's coordinates: $B = \begin{pmatrix} 13 + 2\mu \\ 8 + 2\mu \\ 1 - \mu \end{pmatrix}$

Next, we apply the condition that ( \angle BAP = 45^\circ ). Using the cosine of the angle:

$\cos 45^\circ = \frac{\vec{AB} \cdot \vec{AP}}{|\vec{AB}||\vec{AP}|}$

Calculating ( \vec{AB} ) and ( \vec{AP} ): $\vec{AB} = B - A = \begin{pmatrix} (13 + 2\mu) - 3 \\ (8 + 2\mu) + 2 \\ (1 - \mu) - 6 \end{pmatrix} = \begin{pmatrix} 10 + 2\mu \\ 10 + 2\mu \\ -5 - \mu \end{pmatrix}$

And for ( \vec{AP} ): $\vec{AP} = P - A = \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix} - \begin{pmatrix} 3 \\ -2 \\ 6 \end{pmatrix} = \begin{pmatrix} -4 \\ 2 \\ -6 \end{pmatrix}$

Setting up the equation:

$\cos 45^\circ = \frac{(10 + 2\mu)(-4) + (10 + 2\mu)(2) + (-5 - \mu)(-6)}{\sqrt{(10 + 2\mu)^2 + (10 + 2\mu)^2 + (-5 - \mu)^2} \cdot \sqrt{(-4)^2 + (2)^2 + (-6)^2}}$

This leads to two possible solutions for B's coordinates, and calculating this yields two sets of answers after solving the equations.

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 3 - 2013 - Paper 9

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 3 - 2013 - Paper 9

find the value of p.

find the coordinates of the two possible positions of B.

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