Question

A hunter is at (4, -1, 5) units. He observes two preys at P₁(-1, 2, 0) units and P₂(1, 1, 4), respectively. At zero instant he starts moving in the plane of their positions with uniform speed of 5 units s^-1s in a direction perpendicular to line P₁P₂ till he sees P₁ and P₂ collinear at time T. Time T is

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

Answer 1

Answer: (A)

0.53 s

Answer 2

Here $\overrightarrow{A}$ = (- 1 - 4)$\widehat{i}$ + (2 + 1)$\widehat{j}$ + (0 - 5)$\widehat{k}$ = -5$\widehat{i}$ + 3$\widehat{j}$ - 5$\widehat{j}$ and $\overrightarrow{B}$ = (1 + 1)$\widehat{i}$ + (1 - 2)$\widehat{j}$ + (4 - 0)$\widehat{k}$ = 2$\widehat{i}$ - $\widehat{j}$ + 4$\widehat{k}$

Now R = |$\overrightarrow{A}$| sin θ and |$\overrightarrow{A}$ + $\overrightarrow{B}$| = AB sin θ or sin θ = $\frac{|\overrightarrow{A} \times \overrightarrow{B}|}{AB}i,e.,R = A\frac{|\overrightarrow{A} \times \overrightarrow{B}|}{AB} = \frac{|\overrightarrow{A} \times \overrightarrow{B}|}{B}$

Now $\overrightarrow{A} \times \overrightarrow{B}$ = $\left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \

• 5 & 3 & - 5 \ 2 & - 1 & 4 \end{matrix} \right|$

= (12 - 5)$\widehat{i}$ + (-10 + 20)$\widehat{j}$ + (5 - 6)$\widehat{k}$ = 7$\widehat{i}$ + 10$\widehat{j}$ - $\widehat{k}$

|$\overrightarrow{A}$ × $\overrightarrow{B}$| = (7² + 10² + 1)^1/2 = 12.25 |$\overrightarrow{B}$|

= (2² + 1² + 4²)^1/2 = 4.58

∴ R = 12.25/4.58 = 2.67 m

Time taken to reach destination

= R/v = 2.67/5 = 0.53 s