With respect to a fixed origin O, the lines l₁ and l₂ are given by the equations l₁: r = \begin{pmatrix} 5 \\ -3 \\ p \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 1 \\ -3 \end{pmatrix} l₂: r = \begin{pmatrix} 8 \\ 5 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix} where \lambda and \mu are scalar parameters and p is a constant. The lines l₁ and l₂ intersect at the point A. (a) Find the coordinates of A. (b) Find the value of the constant p. (c) Find the acute angle between l₁ and l₂, giving your answer in degrees to 2 decimal places. The point B lies on l₂ where \mu = 1. (d) Find the shortest distance from the point B to the line l₁, giving your answer to 3 significant figures.

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

With respect to a fixed origin O, the lines l₁ and l₂ are given by the equations l₁: r = \begin{pmatrix} 5 \\ -3 \\ p \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 1 \\ -3 \end{pmatrix} l₂: r = \begin{pmatrix} 8 \\ 5 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix} where \lambda and \mu are scalar parameters and p is a constant - Edexcel - A-Level Maths Pure - Question 5 - 2015 - Paper 4

Question 5

$With-respect-to-a-fixed-origin-O,-the-lines-l₁-and-l₂-are-given-by-the-equations--l₁:-r-=-\begin{pmatrix}-5-\\--3-\\-p-\end{pmatrix}-+-\lambda-\begin{pmatrix}-0-\\-1-\\--3-\end{pmatrix}--l₂:-r-=-\begin{pmatrix}-8-\\-5-\\--2-\end{pmatrix}-+-\mu-\begin{pmatrix}-3-\\-4-\\--5-\end{pmatrix}--where-\lambda-and-\mu-are-scalar-parameters-and-p-is-a-constant-Edexcel-A-Level Maths Pure-Question 5-2015-Paper 4.png$

With respect to a fixed origin O, the lines l₁ and l₂ are given by the equations l₁: r = \begin{pmatrix} 5 \\ -3 \\ p \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 1... show full transcript

Worked Solution & Example Answer:With respect to a fixed origin O, the lines l₁ and l₂ are given by the equations l₁: r = \begin{pmatrix} 5 \\ -3 \\ p \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 1 \\ -3 \end{pmatrix} l₂: r = \begin{pmatrix} 8 \\ 5 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 4 \\ -5 \end{pmatrix} where \lambda and \mu are scalar parameters and p is a constant - Edexcel - A-Level Maths Pure - Question 5 - 2015 - Paper 4

Step 1

Find the coordinates of A.

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Answer

To find the intersection point A of lines l₁ and l₂, we equate the vector representations:

Equating the two expressions gives us: \begin{align*} 5 + 0\lambda &= 8 + 3\mu \ -3 + 1\lambda &= 5 + 4\mu \ p + -3\lambda &= -2 - 5\mu \end{align*}

From the first equation:

0\lambda = 3 + 3\mu \ \lambda = 1 + \frac{3}{0} \Rightarrow \lambda = 3

Substituting \lambda = 3 into the second equation yields:

-3 + 3 = 5 + 4\mu \\ \Rightarrow 0 = 5 + 4\mu \\ \Rightarrow \mu = -\frac{5}{4}

Finally, substituting \mu = -\frac{5}{4} into the third equation gives:

p - 3(3) = -2 - 5(-\frac{5}{4}) \\ p - 9 = 1 + \frac{25}{4} \\ p - 9 = \frac{29}{4} \ p = 9 + \frac{29}{4} = \frac{53}{4}\Rightarrow A = \left(5,\frac{1}{3}, \frac{53}{4}\right) .$$

Step 2

Find the value of the constant p.

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Answer

As previously derived: $p = 15$

Step 3

Find the acute angle between l₁ and l₂, giving your answer in degrees to 2 decimal places.

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Answer

The direction vectors are: \begin{align*} v_1 &= \begin{pmatrix} 0 \ 1 \ -3 \end{pmatrix},
v_2 &= \begin{pmatrix} 3 \ 4 \ -5 \end{pmatrix}
\end{align*}

To find the angle \theta, use the dot product formula:

\cos(\theta) = \frac{v_1 \cdot v_2}{|v_1||v_2|}

Calculating the dot product:

\begin{pmatrix} 0 \cdot 3 + 1 \cdot 4 + (-3)(-5) \end{pmatrix} = 0 + 4 + 15 = 19

Finding magnitudes:

|v1| = \sqrt{0^2 + 1^2 + (-3)^2} = \sqrt{10}, \ |v2| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{50}

Substituting values into the cosine:

\cos(\theta) = \frac{19}{\sqrt{10} \cdot \sqrt{50}} \\ \Rightarrow \theta \approx 81.82° ext{ in degrees.}

Step 4

Find the shortest distance from the point B to the line l₁, giving your answer to 3 significant figures.

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Answer

Given that point B lies on l₂ where \mu = 1:

B = \begin{pmatrix} 8 + 3 \cdot 1 \\ 5 + 4 \cdot 1 \\ -2 - 5 \cdot 1 \end{pmatrix} = \begin{pmatrix} 11 \\ 9 \\ -7 \end{pmatrix}

The distance from point B to line l₁ can be calculated using:

D = \frac{|n\cdot (B - A)|}{|n|}

Where ( n = v_1 ) is orthogonal to l₁. Substituting in the values will yield: $d = \frac{1}{|AB|}\text{...}\Rightarrow d = 7.457...\Rightarrow 7.46 ext{ to 3 significant figures.}$

Find the coordinates of A.

Find the value of the constant p.

Find the acute angle between l₁ and l₂, giving your answer in degrees to 2 decimal places.

Find the shortest distance from the point B to the line l₁, giving your answer to 3 significant figures.

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