All coordinates in this question are in metres. In a warehouse where a crime involving the use of fi

Question

All coordinates in this question are in metres. In a warehouse where a crime involving the use of firearms occurred, an eyewitness claimed to have seen the shooter fire $2$ shots. The first bullet hit a box and made $2$ holes at A and B. The second bullet hit the victim and broke the window. According to the eyewitness account, he was at the door and the shooter was somewhere near the centre of the room. A schematic of the room is as shown in Figure 1 below, with the approximate standing positions of the eyewitness, shooter and victim marked by crosses. A crime scene investigator wished to determine the location from which the shot was fired by reconstructing the path of the shot that made the $2$ bullet holes in the box. The investigator assigned the coordinates of the approximate standing positions of the eyewitness, the shooter and the victim as $\left( 0,0,0 \right)$, $\left( -2,5,0 \right)$ and $\left( 0,9,0 \right)$ respectively. The investigator also assigned the coordinates of the positions of the $2$ bullet holes at A and B as $\left( 3.42,\,\,8.72,\,\,0.80 \right)$ and $\left( 4.35,\,\,9.32,\,\,0.70 \right)$ respectively. Assume the bullets travelled in straight lines. Find a cartesian equation of the path of the first shot passing through both bullet holes.[2] By considering the standing position of the shooter and the distance between the weapon and the ground, explain why the investigator believed that the evidence of the $2$ bullet holes supported the eyewitness account.[3] After piecing back the broken window, the investigator assigned the coordinates of the position of the impact as $\left( 0.54,\,\,9.97,\,\,1.50 \right)$. The bullet wound on the victim was on his right shoulder, at a height of $1.48$m from the floor. Find a vector equation of the path of the second shot hitting the victim and the window.[2] Using the results in parts (i) and (iii), find the coordinates of the standing position of the shooter.[3] State two other assumptions made in the above calculations.[2]

Accepted Answer

(i) $\frac{x-3.42}{0.93}=\frac{y-8.72}{0.6}=\frac{z-0.8}{-0.1}$ (ii) When $x=-2$, $\frac{-2-3.42}{0.93}=\frac{y-8.72}{0.6}=\frac{z-0.8}{-0.1}$,$y=5.223$, $z=1.383$ OR when $y=5$, $\frac{x-3.42}{0.93}=\frac{5-8.72}{0.6}=\frac{z-0.8}{-0.1}$, $x=-2.346$, $z=1.42$. As the location of the shooter is approximate, the difference of $22.3$ cm (or $34.6$ cm) from $\left( -2,5 \right)$ is acceptable. (A person’s elbow to gun tip easily exceeds $30$ cm). Also, $1.38$ m (or $1.42$ m) is a reasonable height for the tip of the weapon to be at when held by a standing person. (Depending on the height of the person.) (iii) ${{l}_{VW}}:\mathbf{r}=\left( \begin{matrix} 0 \ 9 \ 1.48 \ \end{matrix} \right)+\mu \left( \begin{matrix} 0.54 \ 0.97 \ 0.02 \ \end{matrix} \right),\mu \in \mathbb{R}$ (iv) $\left( -2.16,\,\,5.12,\,\,0 \right)$ (v) The shooter remains at the same position for both sides. The box remains at the same position after the first impact of hole A. The bullets continue to travel in the same straight line after hitting the first object.

2018 HCI J2 CT1 Q10

2018 HCI J2 CT1 Q10

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