The line $l_1$ has equation $$ r = \begin{pmatrix} 2 \\\ 3 \\\ -4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\\ 2 \\\ 1 \end{pmatrix},$$ where $\lambda$ is a scalar parameter. The line $l_2$ has equation $$ r = \begin{pmatrix} 0 \\\ 9 \\\ 0 \end{pmatrix} + \mu \begin{pmatrix} 5 \\\ 0 \\\ 2 \end{pmatrix},$$ where $\mu$ is a scalar parameter. Given that $l_1$ and $l_2$ meet at the point $C$, find a) the coordinates of $C$. The point $A$ is the point on $l_1$ where $\lambda = 0$ and the point $B$ is the point on $l_1$, where $\mu = -1$. b) Find the size of the angle $ACB$. Give your answer in degrees to 2 decimal places. c) Hence, or otherwise, find the area of the triangle $ABC$.

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The line $l_1$ has equation $$ r = \begin{pmatrix} 2 \\\ 3 \\\ -4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\\ 2 \\\ 1 \end{pmatrix},$$ where $\lambda$ is a scalar parameter - Edexcel - A-Level Maths Pure - Question 1 - 2010 - Paper 6

Question 1

$The-line-$l_1$-has-equation-$$-r-=-\begin{pmatrix}-2-\\\-3-\\\--4-\end{pmatrix}-+-\lambda-\begin{pmatrix}-1-\\\-2-\\\-1-\end{pmatrix},$$-where-$\lambda$-is-a-scalar-parameter-Edexcel-A-Level Maths Pure-Question 1-2010-Paper 6.png$

The line $l_1$ has equation $$ r = \begin{pmatrix} 2 \\\ 3 \\\ -4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\\ 2 \\\ 1 \end{pmatrix},$$ where $\lambda$ is a scalar ... show full transcript

Worked Solution & Example Answer:The line $l_1$ has equation $$ r = \begin{pmatrix} 2 \\\ 3 \\\ -4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\\ 2 \\\ 1 \end{pmatrix},$$ where $\lambda$ is a scalar parameter - Edexcel - A-Level Maths Pure - Question 1 - 2010 - Paper 6

Step 1

a) the coordinates of C.

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Answer

To find the coordinates of the point $C$ , we need to set the two equations for $l_1$ and $l_2$ equal to each other:

From $l_1$ : $r_1 = \begin{pmatrix} 2 \ + \lambda \cdot 1 \\\ 3 \ + \lambda \cdot 2 \\\ -4 \ + \lambda \cdot 1 \end{pmatrix} = \begin{pmatrix} 2 + \lambda \\\ 3 + 2\lambda \\\ -4 + \lambda \end{pmatrix}$

From $l_2$ : $r_2 = \begin{pmatrix} 0 \ + \mu \cdot 5 \\\ 9 \ + \mu \cdot 0 \\\ 0 \ + \mu \cdot 2 \end{pmatrix} = \begin{pmatrix} 5\mu \\\ 9 \\\ 2\mu \end{pmatrix}$

Setting the corresponding components equal to each other:

$2 + \lambda = 5\mu$
$3 + 2\lambda = 9$
$-4 + \lambda = 2\mu$

From the second equation, we can solve for \lambda: $3 + 2\lambda = 9 \Rightarrow 2\lambda = 6 \Rightarrow \lambda = 3$ Substituting \lambda back into the first equation: $2 + 3 = 5\mu \Rightarrow 5 = 5\mu \Rightarrow \mu = 1$

Now using \lambda = 3 in one of the equations to find the coordinates of $C$ :

Using \lambda = 3 in $l_1$ : $C = \begin{pmatrix} 2 + 3 \\\ 3 + 2 \cdot 3 \\\ -4 + 3 \end{pmatrix} = \begin{pmatrix} 5 \\\ 9 \\\ -1 \end{pmatrix}$

Thus, the coordinates of point $C$ are $(5, 9, -1)$ .

Step 2

b) Find the size of the angle ACB.

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Answer

To find the angle $ACB$ , we first need the direction vectors $\overrightarrow{AC}$ and $\overrightarrow{BC}$ .

The coordinates of points:

$A = (2, 3, -4)$ ,
$B = (-5, -9, -5)$ ,
$C = (5, 9, -1)$ .

Calculate vectors:

$\overrightarrow{AC} = C - A = \begin{pmatrix} 5 - 2 \\\ 9 - 3 \\\ -1 - (-4) \end{pmatrix} = \begin{pmatrix} 3 \\\ 6 \\\ 3 \end{pmatrix}$
$\overrightarrow{BC} = C - B = \begin{pmatrix} 5 - (-5) \\\ 9 - (-9) \\\ -1 - (-5) \end{pmatrix} = \begin{pmatrix} 10 \\\ 18 \\\ 4 \end{pmatrix}$

Now, find the dot product: $\overrightarrow{AC} \cdot \overrightarrow{BC} = (3)(10) + (6)(18) + (3)(4) = 30 + 108 + 12 = 150$

Find magnitudes: $|\overrightarrow{AC}| = \sqrt{3^2 + 6^2 + 3^2} = \sqrt{9 + 36 + 9} = \sqrt{54} = 3\sqrt{6}$ $|\overrightarrow{BC}| = \sqrt{10^2 + 18^2 + 4^2} = \sqrt{100 + 324 + 16} = \sqrt{440} = 2\sqrt{110}$

Use the cosine formula for angle $\angle ACB$ : $\cos \theta = \frac{\overrightarrow{AC} \cdot \overrightarrow{BC}}{|\overrightarrow{AC}| |\overrightarrow{BC}|} = \frac{150}{(3\sqrt{6})(2\sqrt{110})}$ Calculating gives: $\theta = \cos^{-1} \left( \frac{150}{6\sqrt{660}}\right) \approx 57.95 \text{ degrees.}$

Step 3

c) Hence, or otherwise, find the area of the triangle ABC.

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Answer

To find the area of triangle $ABC$ , we can use the formula: $\text{Area} = \frac{1}{2} |\overrightarrow{AC} \times \overrightarrow{BC}|$

First, let's compute the cross product $\overrightarrow{AC} \times \overrightarrow{BC}$ .

Using components:

$\overrightarrow{AC} = \begin{pmatrix} 3 \\\ 6 \\\ 3 \end{pmatrix}$
$\overrightarrow{BC} = \begin{pmatrix} 10 \\\ 18 \\\ 4 \end{pmatrix}$

Cross product: $\overrightarrow{AC} \times \overrightarrow{BC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ 3 & 6 & 3 \\\ 10 & 18 & 4 \end{vmatrix}$

Calculating the determinant gives: $\hat{i}(6 \cdot 4 - 3 \cdot 18) - \hat{j}(3 \cdot 4 - 3 \cdot 10) + \hat{k}(3 \cdot 18 - 6 \cdot 10)$ $= \hat{i}(24 - 54) - \hat{j}(12 - 30) + \hat{k}(54 - 60)$ $= -30\hat{i} + 18\hat{j} - 6\hat{k}$

Magnitude of the cross product: $|\overrightarrow{AC} \times \overrightarrow{BC}| = \sqrt{(-30)^2 + 18^2 + (-6)^2} = \sqrt{900 + 324 + 36} = \sqrt{1260} = 6\sqrt{35}$

Thus, the area of triangle $ABC$ is: $\text{Area} = \frac{1}{2} |\overrightarrow{AC} \times \overrightarrow{BC}| = \frac{1}{2}(6\sqrt{35}) \approx 15\sqrt{5} \approx 33.5$

The line $l_1$ has equation $$ r = \begin{pmatrix} 2 \\\ 3 \\\ -4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\\ 2 \\\ 1 \end{pmatrix},$$ where $\lambda$ is a scalar parameter - Edexcel - A-Level Maths Pure - Question 1 - 2010 - Paper 6

Worked Solution & Example Answer:The line $l_1$ has equation $$ r = \begin{pmatrix} 2 \\\ 3 \\\ -4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\\ 2 \\\ 1 \end{pmatrix},$$ where $\lambda$ is a scalar parameter - Edexcel - A-Level Maths Pure - Question 1 - 2010 - Paper 6

a) the coordinates of C.

b) Find the size of the angle ACB.

c) Hence, or otherwise, find the area of the triangle ABC.

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