The line $l_1$ has equation $$ extbf{r} = \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$$ where $\\lambda$ is a scalar parameter. The line $l_2$ has equation $$ extbf{r} = \begin{pmatrix} 0 \\ 9 \\ 0 \end{pmatrix} + \\mu \begin{pmatrix} 5 \\ 0 \\ 2 \end{pmatrix}$$ 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 $$ extbf{r} = \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$$ where $\\lambda$ is a scalar parameter - Edexcel - A-Level Maths Pure - Question 1 - 2009 - Paper 6

Question 1

$The-line-$l_1$-has-equation--$$-extbf{r}-=-\begin{pmatrix}-2-\\-3-\\--4-\end{pmatrix}-+-\lambda-\begin{pmatrix}-1-\\-2-\\-1-\end{pmatrix}$$-where-$\\lambda$-is-a-scalar-parameter-Edexcel-A-Level Maths Pure-Question 1-2009-Paper 6.png$

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

Worked Solution & Example Answer:The line $l_1$ has equation $$ extbf{r} = \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$$ where $\\lambda$ is a scalar parameter - Edexcel - A-Level Maths Pure - Question 1 - 2009 - Paper 6

Step 1

Find the coordinates of C

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Answer

To find the coordinates of point $C$ , we set up the following equations from the two lines:

From $l_1$ : $\begin{pmatrix} 2 + \lambda \\ 3 + 2\lambda \\ -4 + \lambda \end{pmatrix}$ From $l_2$ : $\begin{pmatrix} 5\mu \\ 9 \\ 2\mu \end{pmatrix}$

Equating the components, we get the following system of equations:

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

From the second equation, we can solve for $\lambda$ : $2\lambda = 9 - 3 \\ \lambda = 3$

Substituting $\lambda = 3$ into the first equation: $2 + 3 = 5\mu \\ 5 = 5\mu \\ \mu = 1$

Now substituting $\lambda = 3$ into the line $l_1$ gives us the coordinates of point C: $$\textbf{C} = \begin{pmatrix} 2 + 3 \ 3 + 2(3) \ -4 + 3 \end{pmatrix} = \begin{pmatrix} 5 \ 9 \ -1 \end{pmatrix}.$

Step 2

Find the size of the angle ACB

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Answer

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

Point $A$ has coordinates when $\lambda = 0$ :
$\textbf{A} = \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix}$
Point $B$ has coordinates when $\mu = -1$ : $\textbf{B} = \begin{pmatrix} 0 + 5(-1) \\ 9 \\ 0 + 2(-1) \end{pmatrix} = \begin{pmatrix} -5 \\ 9 \\ -2 \end{pmatrix}$

Now, we can find the vectors:

$\overrightarrow{AC} = \textbf{C} - \textbf{A} = \begin{pmatrix} 5 \ - 2 \ - 1 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \\ 6 \\ 3 \end{pmatrix}$

$\overrightarrow{BC} = \textbf{C} - \textbf{B} = \begin{pmatrix} 5 \\ 9 \\ -1 \end{pmatrix} - \begin{pmatrix} -5 \\ 9 \\ -2 \end{pmatrix} = \begin{pmatrix} 10 \\ 0 \\ 1 \end{pmatrix}$

Next, we calculate the cosine of the angle $\theta$ between $\overrightarrow{AC}$ and $\overrightarrow{BC}$ using the dot product:

$\cos \theta = \frac{\overrightarrow{AC} \cdot \overrightarrow{BC}}{|\overrightarrow{AC}||\overrightarrow{BC}|}$

Calculating the dot product: $\overrightarrow{AC} \cdot \overrightarrow{BC} = 3 \cdot 10 + 6 \cdot 0 + 3 \cdot 1 = 30 + 0 + 3 = 33$

Now calculate the magnitudes: $|\overrightarrow{AC}| = \sqrt{3^2 + 6^2 + 3^2} = \sqrt{9 + 36 + 9} = \sqrt{54}$

$|\overrightarrow{BC}| = \sqrt{10^2 + 0^2 + 1^2} = \sqrt{100 + 0 + 1} = \sqrt{101}$

Thus, $\cos \theta = \frac{33}{\sqrt{54} \cdot \sqrt{101}}.$

Calculating $\theta$ gives: $$\theta = \cos^{-1} \left(\frac{33}{\sqrt{54 \cdot 101}}\right) \approx 57.95^\circ.$

Step 3

Hence, or otherwise, find the area of the triangle ABC

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Answer

The area $A$ of triangle $ABC$ can be calculated using the formula:

$A = \frac{1}{2} \cdot |\overrightarrow{AC}| \cdot |\overrightarrow{BC}| \cdot \sin(\theta)$

We already have: $|\overrightarrow{AC}| = \sqrt{54} = 3\sqrt{6}$ $|\overrightarrow{BC}| = \sqrt{101}$ $\theta \approx 57.95^\circ$

Thus, $A = \frac{1}{2} \cdot (3\sqrt{6}) \cdot \sqrt{101} \cdot \sin(57.95^\circ)$

Using the sine value: $\sin(57.95^\circ) \approx 0.8387$

Finally, compute: $A = \frac{1}{2} \cdot 3\sqrt{6} \cdot \sqrt{101} \cdot 0.8387 \approx 33.5.$

The line $l_1$ has equation $$ extbf{r} = \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$$ where $\\lambda$ is a scalar parameter - Edexcel - A-Level Maths Pure - Question 1 - 2009 - Paper 6

Worked Solution & Example Answer:The line $l_1$ has equation $$ extbf{r} = \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$$ where $\\lambda$ is a scalar parameter - Edexcel - A-Level Maths Pure - Question 1 - 2009 - Paper 6

Find the coordinates of C

Find the size of the angle ACB

Hence, or otherwise, find the area of the triangle ABC

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