| Exam Board | SPS |
|---|---|
| Module | SPS FM (SPS FM) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Topic | Vectors: Cross Product & Distances |
| Type | Area of triangle using cross product |
| Difficulty | Standard +0.8 This is a multi-step further maths vectors question requiring: (1) converting Cartesian line equations to parametric form, (2) solving simultaneous equations to find intersection point C, (3) forming vectors AC and BC, and (4) using cross product to find triangle area. While the individual techniques are standard for FM, the combination of steps and the need to work carefully with two different line formats makes this moderately challenging. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting4.04g Vector product: a x b perpendicular vector |
12. The line $l _ { 1 }$ has Cartesian equation
$$x - 2 = \frac { y - 3 } { 2 } = z + 4$$
The line $l _ { 2 }$ has Cartesian equation
$$\frac { x } { 5 } = \frac { z + 3 } { 2 } , \quad y = 9$$
Given that $l _ { 1 }$ and $l _ { 2 }$ meet at the point C , find
\begin{enumerate}[label=(\alph*)]
\item the coordinates of C .
The point $\mathrm { A } ( 2,3 , - 4 )$ is on the line $l _ { 1 }$ and the point $\mathrm { B } ( - 5,9 , - 5 )$ is on the line $l _ { 2 }$.
\item find the area of the triangle $A B C$.
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM 2020 Q12 [6]}}