Solved: Two distinct lines, ${l_1}$ and ${l_2}$, intersect at a point ${\text{Q}}$. In... [IB Mathematics Analysis and Approaches (AA)]

Question

HLPaper 1

Two distinct lines, ${l_1}$ and ${l_2}$ , intersect at a point ${\text{Q}}$ .

In addition to $Q$ , four distinct points are marked out on ${l_1}$ and three distinct points on ${l_2}$ . A mathematician decides to join some of these eight points to form polygons.

The line $l_1$ has vector equation $l_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$ , ${\lambda \in \mathbb{R}}$

The line $l_2$ has vector equation $l_2 = \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} + \mu \begin{pmatrix} 5 \\ 6 \\ 2 \end{pmatrix}$ , ${\mu \in \mathbb{R}}$ .

The point ${\text{Q}}$ has coordinates (4, 6, 4).

2 extra points given are

The point ${\text{C}}$ has coordinates (3, 4, 3) and lies on ${l_1}$ .

The point ${\text{D}}$ has coordinates (-1, 0, 2) and lies on ${l_2}$ .

Write down the value of ${\lambda}$ corresponding to the point ${\text{C}}$ .

[1]

Verified

Solution

$\begin{pmatrix} 3 \\ 4 \\ 3 \end{pmatrix}$ = $\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$ Hence $\lambda +1 = 3$ hence $\lambda =2$ A1

No need to check if other equations hold at 2 as well because it is given that point $C$ is indeed on $l_1$

Write down $\overrightarrow {{QC}}$ and $\overrightarrow {{QD}}$ .

[2]

Verified

Solution

$\overrightarrow{{QC}} = \begin{pmatrix} -1 \\ -2 \\ -1 \end{pmatrix}, \overrightarrow{{QD}} = \begin{pmatrix} -5 \\ -6 \\ -2 \end{pmatrix}$ A1A1

Note: Award A1A0 if both are given as coordinates.

Given that the polygon has to be made up of the distinct points marked on the lines, find how many sets of four points can be selected which can form the vertices of a quadrilateral.

Each side of the quadrilateral should have some angular deviation from each other.

[4]

Verified

Solution

$Q$ cannot be a point because it will always make triangle M1
We cannot choose more than two points from each side otherwise we will have 2 sides on a straight line (no angular deviation) M1
Hence we need to pick 2 points from each side such that

\begin{pmatrix}4\\2\end{pmatrix} \times \begin{pmatrix}3\\2\end{pmatrix}= 6\times3=18

A1 A1

Find how many sets of three points can be selected which can form the vertices of a triangle.

[4]

Verified

Solution

We cannot choose more than two points from each side otherwise we will have 2 sides on a straight line
If $Q$ is included then we need to take one point from each such that we have

4\times 3 = 12

If $Q$ is not included then we have 2 points from $l_1$ and 1 from $l_2$ or 1 from $l_1$ and 2 from $l_2$

\begin{pmatrix} 4 \\ 2 \end{pmatrix} \times \begin{pmatrix} 3 \\ 1 \end{pmatrix} + \begin{pmatrix} 4 \\ 1 \end{pmatrix} \times \begin{pmatrix} 3 \\ 2 \end{pmatrix} = 6\times3+4\times3=30

A1 M1

Hence total is $42$ possibilities A1

Verify that $Q$ is the point of intersection of the two lines.

[3]

Verified

Solution

$\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} + \mu \begin{pmatrix} 5 \\ 6 \\ 2 \end{pmatrix}$ hence we get

$1+\lambda = 5\mu-1$

$2\lambda = 6\mu$

$1 + \lambda = 2 + 2\mu$

Therefore $\lambda = 3\mu$ and so $1+3\mu = 2+2\mu \Rightarrow \mu=1\Rightarrow\lambda=3$

And so we check if it holds for the first

$1+3=4=5-1=5(1)-1$ Hence it holds so indeed true

Let $E$ be the point on $l_1$ with coordinates (1, 0, 1) and $F$ be the point on $l_2$ with parameter $\mu = -2$ .

Find the area of the quadrilateral $FEDC$ .

[8]

Verified

Solution

METHOD 1

Area triangle $CDQ = \frac{1}{2}|\mathbf{QD} \times \mathbf{QC}|$

$(= \frac{1}{2}|\begin{pmatrix} -5\\-6\\-2\end{pmatrix} \times \begin{pmatrix} -1\\-2\\-1\end{pmatrix}| = \frac{1}{2}|\begin{pmatrix} 2\\-3\\4 \end{pmatrix}|$

$= \frac{\sqrt{29}}{2}$

Either

$\mathbf{QE} = 3\mathbf{QC}$ , $\mathbf{QF} = 3\mathbf{QD}$ (M1)

Area triangle $QEF = 9 \times$ Area triangle $CDQ$ (M1A1)

$= \frac{9\sqrt{29}}{2}$

F has coordinates (−11, −12, −2) A1

Area triangle $QEF = \frac{1}{2}|\mathbf{QF} \times \mathbf{QE}|$

$= \frac{1}{2}|\begin{pmatrix} -15\\-18\\-6 \end{pmatrix} \times \begin{pmatrix} -3\\-6\\-3\end{pmatrix}|$

$= \frac{9\sqrt{29}}{2}$

M1A1

THEN

Area of $FEDC = \frac{9\sqrt{29}}{2} - \frac{\sqrt{29}}{2}$

$= 4\sqrt{29}$

Method 2

F has coordinates (−11,−12,−2) A1

area= $\frac{1}{2}|\mathbf{EC} \times \mathbf{ED}| + \frac{1}{2}|\mathbf{DE} \times \mathbf{DF}|$ ** M1**

$\mathbf{EC} = \begin{pmatrix} - 2\\ 0 \\1 \end{pmatrix}$

$\mathbf{ED} =\begin{pmatrix}2\\4\\2 \end{pmatrix}$

$\mathbf{EC} \times \mathbf{ED} = \begin{pmatrix} -4\\6\\-8 \end{pmatrix}$

$\mathbf{DE} = \begin{pmatrix} 2\\0\\-1 \end{pmatrix}$

$\mathbf{DF} = \begin{pmatrix} -10\\-12\\-4\end{pmatrix}$

$\mathbf{DE} \times \mathbf{DF} = \begin{pmatrix} -12\\18\\-24 \end{pmatrix}$

Area = $\frac{1}{2} \times 2\sqrt{29} + \frac{1}{2} \times 6\sqrt{29}$

$= 4\sqrt{29}$ A1

Two distinct lines, ${l_1}$ and ${l_2}$, intersect at a point ${\text{Q}}$.

Method 2

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