Question

A bird is at a point $P(4m,-1m,5m)$ and sees two points $P_1(-1m,-1m,0m)$ and $P_2(3m,-1m,-3m)$. At time $t=0$, it starts flying in a plane of the three positions, with a constant speed of $2m/s$ in a direction perpendicular to the straight line $P_1{P}_2$ till it sees $P_1\&P_2$ collinear at time $t$. Find the time $t$.

• A. $3.5s$
• B. $2.5s$
• C. $3.4s$
• D. $1.5s$

Answer 1

The correct option is A $3.5s$

Vector $\overrightarrow{P{P}_1}=-5\hat{i}-5\hat{k}$ and $\overrightarrow{P_1{P}_2}=4\hat{i}-3\hat{k}$

Let angle between these vectors be $\theta$ then

$cos\theta =\frac{\left(\begin{array}{c}-5\hat{i}-5\hat{k}\end{array}\right)\cdot \left(\begin{array}{c}4\hat{i}-3\hat{k}\end{array}\right)}{\left(\begin{array}{c}5\sqrt{2}\end{array}\right)\left(5\right)}=-\frac{1}{5\sqrt{2}}$

As $PM=P{P}_{1}sin\theta$

So $PM=\left(5\sqrt{2}\right)\left(\begin{array}{c}\frac{7}{5\sqrt{2}}\end{array}\right)=7m$

Therefore $t=\frac{7m}{2m/s}=3.5s$