Question

A hunter is at $(4,-1,5)$ units. He observes two preys at $P_1(-1,2,0)$ units and $P_2(1,1,4)$ respectively. At zero instant he starts moving in the plane of their positions with uniform speed of $5units{s}^{-1}$ in a direction perpendicular to line $P_1{P}_2$ till he sees $P_1$ and $P_2$ collinear at time $T$. Time $T$ is

• A. $0.53s$
• B. $0.73s$
• C. $0.35s$
• D. $0.92s$

Answer 1

The correct option is A $0.53s$

Let,
A be the vector acting along hunter and $P_1$ and
B be the vector acting along $P_1$ and $P_2$
Also,
R be the vector of velocity with which hunter travels perpendicular to $P_1.P_2$
Then,
$\overrightarrow{A}=(-1-4)\hat{i}+(2+1)\hat{j}+(0-5)\hat{k}=-5\hat{i}+3\hat{j}-5\hat{k}$ and
$\overrightarrow{B}=(1+1)\hat{i}+(1-2)\hat{j}+(4-0)\hat{k}=2\hat{i}-\hat{j}+4\hat{k}$
Now,
$R=|\overrightarrow{A}|\sin \theta$ and
$|\overrightarrow{A}\times \overrightarrow{B}|=AB\sin \theta$ or
$\sin \theta =\frac{|\overrightarrow{A}\times \overrightarrow{B}|}{AB}$
i.e $R=A\frac{|\overrightarrow{A}\times \overrightarrow{B}|}{AB}=\frac{|\overrightarrow{A}\times \overrightarrow{B}|}{B}$
Now, $\overrightarrow{A}\times \overrightarrow{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -5& 3& -5\\ 2& -1& 4\end{array}\right|$
$=(12-5)\hat{i}+(-10+20)\hat{j}+(5-6)\hat{k}=7\hat{i}+10\hat{j}-\hat{k}$
$|\overrightarrow{A}\times \overrightarrow{B}|=(7^2+{10}^2+1{)}^{1/2}=12.25$
$|\overrightarrow{B}|=(2^2+1^2+4^2{)}^{1/2}=4.58$
$\therefore R=\frac{12.25}{4.58}=2.67m$
Time taken to reach destination,
$=\frac{R}{v}=\frac{2.67}{5}=0.53s$