Question: Two distinct lines, $l_1$ and $l_2$, intersect at a point $P$. In addition to $P$, 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.
(a) (i) Find how many sets of four points can be selected which can form the vertices of a quadrilateral.
(ii) Find how many sets of three points can be selected which can form the vertices of a triangle.
The line $l_1$ has vector equation $$\mathbf{r}_1 = \begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix} + \lambda \begin{pmatrix}1 \\ 2 \\ 1\end{pmatrix}, \lambda \in \mathbb{R}$$ and the line $l_2$ has vector equation
$$\mathbf{r}_2 = \begin{pmatrix}-1 \\ 0 \\ 2\end{pmatrix} + \mu \begin{pmatrix}5 \\ 6 \\ 2\end{pmatrix}, \mu \in \mathbb{R}.$$
The point $P$ has coordinates $(4, 6, 4)$.
(b) Verify that $P$ is the point of intersection of the two lines.
The point $A$ has coordinates $(3, 4, 3)$ and lies on $l_1$.
(c) Write down the value of $\lambda$ corresponding to the point $A$.
The point $B$ has coordinates $(-1, 0, 2)$ and lies on $l_2$.
(d) Write down $\overrightarrow{PA}$ and $\overrightarrow{PB}$.
Let $C$ be the point on $l_1$ with coordinates $(1, 0, 1)$ and $D$ be the point on $l_2$ with parameter $\mu = -2$.
(e) Find the area of the quadrilateral $CDBA$.
1. **Problem Statement:**
We have two intersecting lines $l_1$ and $l_2$ with points marked on them. We want to find:
(a)(i) Number of sets of four points forming quadrilaterals.
(a)(ii) Number of sets of three points forming triangles.
(b) Verify $P=(4,6,4)$ lies on both lines.
(c) Find $\lambda$ for point $A=(3,4,3)$ on $l_1$.
(d) Find vectors $\overrightarrow{PA}$ and $\overrightarrow{PB}$.
(e) Find area of quadrilateral $CDBA$ with $C=(1,0,1)$ on $l_1$ and $D$ on $l_2$ at $\mu=-2$.
2. **(a)(i) Number of quadrilaterals:**
- Total points: $8$ (4 on $l_1$ + 3 on $l_2$ + intersection $P$)
- Quadrilateral vertices are sets of 4 points that do not lie all on the same line.
- Points on $l_1$: $P$ + 4 distinct points = 5 points
- Points on $l_2$: $P$ + 3 distinct points = 4 points
Number of ways to choose 4 points from 8:
$$\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$$
Subtract sets of 4 points all on the same line:
- On $l_1$, 5 points, so $\binom{5}{4} = 5$ sets
- On $l_2$, 4 points, so $\binom{4}{4} = 1$ set
So, number of quadrilaterals:
$$70 - (5 + 1) = 64$$
3. **(a)(ii) Number of triangles:**
- Total points: 8
- Number of ways to choose 3 points:
$$\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$
Subtract sets of 3 points all on the same line (collinear points do not form triangles):
- On $l_1$, 5 points, so $\binom{5}{3} = 10$ sets
- On $l_2$, 4 points, so $\binom{4}{3} = 4$ sets
Number of triangles:
$$56 - (10 + 4) = 42$$
4. **(b) Verify $P$ lies on both lines:**
- For $l_1$, vector equation:
$$\mathbf{r}_1 = \begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix} + \lambda \begin{pmatrix}1 \\ 2 \\ 1\end{pmatrix}$$
Set equal to $P = (4,6,4)$:
$$\begin{cases}1 + \lambda = 4 \\ 0 + 2\lambda = 6 \\ 1 + \lambda = 4\end{cases}$$
From first and third:
$$\lambda = 3$$
From second:
$$2\lambda = 6 \Rightarrow \lambda = 3$$
Consistent, so $P$ lies on $l_1$.
- For $l_2$:
$$\mathbf{r}_2 = \begin{pmatrix}-1 \\ 0 \\ 2\end{pmatrix} + \mu \begin{pmatrix}5 \\ 6 \\ 2\end{pmatrix}$$
Set equal to $P = (4,6,4)$:
$$\begin{cases}-1 + 5\mu = 4 \\ 0 + 6\mu = 6 \\ 2 + 2\mu = 4\end{cases}$$
From second:
$$6\mu = 6 \Rightarrow \mu = 1$$
Check first:
$$-1 + 5(1) = 4$$
Check third:
$$2 + 2(1) = 4$$
All consistent, so $P$ lies on $l_2$.
5. **(c) Find $\lambda$ for $A=(3,4,3)$ on $l_1$:**
$$\begin{cases}1 + \lambda = 3 \\ 0 + 2\lambda = 4 \\ 1 + \lambda = 3\end{cases}$$
From first and third:
$$\lambda = 2$$
From second:
$$2\lambda = 4 \Rightarrow \lambda = 2$$
Consistent, so $\boxed{\lambda = 2}$.
6. **(d) Find $\overrightarrow{PA}$ and $\overrightarrow{PB}$:**
- $P = (4,6,4)$, $A = (3,4,3)$, $B = (-1,0,2)$
$$\overrightarrow{PA} = A - P = (3-4, 4-6, 3-4) = (-1, -2, -1)$$
$$\overrightarrow{PB} = B - P = (-1-4, 0-6, 2-4) = (-5, -6, -2)$$
7. **(e) Find area of quadrilateral $CDBA$:**
- Points:
- $C = (1,0,1)$ on $l_1$
- $D$ on $l_2$ with $\mu = -2$
Calculate $D$:
$$D = \begin{pmatrix}-1 \\ 0 \\ 2\end{pmatrix} + (-2) \begin{pmatrix}5 \\ 6 \\ 2\end{pmatrix} = \begin{pmatrix}-1 - 10 \\ 0 - 12 \\ 2 - 4\end{pmatrix} = (-11, -12, -2)$$
- Quadrilateral vertices in order: $C, D, B, A$
We find area by dividing into two triangles: $CDB$ and $CBA$.
- Vectors for triangle $CDB$:
$$\overrightarrow{CD} = D - C = (-11 - 1, -12 - 0, -2 - 1) = (-12, -12, -3)$$
$$\overrightarrow{CB} = B - C = (-1 - 1, 0 - 0, 2 - 1) = (-2, 0, 1)$$
Area of triangle $CDB$:
$$\text{Area} = \frac{1}{2} \left| \overrightarrow{CD} \times \overrightarrow{CB} \right|$$
Calculate cross product:
$$\overrightarrow{CD} \times \overrightarrow{CB} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -12 & -12 & -3 \\ -2 & 0 & 1 \end{vmatrix} = \mathbf{i}((-12)(1) - (-3)(0)) - \mathbf{j}((-12)(1) - (-3)(-2)) + \mathbf{k}((-12)(0) - (-12)(-2))$$
$$= \mathbf{i}(-12 - 0) - \mathbf{j}(-12 - 6) + \mathbf{k}(0 - 24) = (-12)\mathbf{i} - (-18)\mathbf{j} - 24\mathbf{k} = (-12, 18, -24)$$
Magnitude:
$$\sqrt{(-12)^2 + 18^2 + (-24)^2} = \sqrt{144 + 324 + 576} = \sqrt{1044} = 2\sqrt{261}$$
Area $CDB$:
$$= \frac{1}{2} \times 2\sqrt{261} = \sqrt{261}$$
- Vectors for triangle $CBA$:
$$\overrightarrow{CB} = (-2, 0, 1)$$
$$\overrightarrow{CA} = A - C = (3 - 1, 4 - 0, 3 - 1) = (2, 4, 2)$$
Cross product:
$$\overrightarrow{CB} \times \overrightarrow{CA} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 0 & 1 \\ 2 & 4 & 2 \end{vmatrix} = \mathbf{i}(0 \times 2 - 1 \times 4) - \mathbf{j}(-2 \times 2 - 1 \times 2) + \mathbf{k}(-2 \times 4 - 0 \times 2)$$
$$= \mathbf{i}(0 - 4) - \mathbf{j}(-4 - 2) + \mathbf{k}(-8 - 0) = (-4, 6, -8)$$
Magnitude:
$$\sqrt{(-4)^2 + 6^2 + (-8)^2} = \sqrt{16 + 36 + 64} = \sqrt{116} = 2\sqrt{29}$$
Area $CBA$:
$$= \frac{1}{2} \times 2\sqrt{29} = \sqrt{29}$$
- Total area of quadrilateral $CDBA$:
$$\sqrt{261} + \sqrt{29}$$
**Final answers:**
- (a)(i) Number of quadrilaterals: $64$
- (a)(ii) Number of triangles: $42$
- (b) $P$ lies on both lines.
- (c) $\lambda = 2$
- (d) $\overrightarrow{PA} = (-1, -2, -1)$, $\overrightarrow{PB} = (-5, -6, -2)$
- (e) Area of quadrilateral $CDBA = \sqrt{261} + \sqrt{29}$