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Multiplication with Vectors

Two persons ...

Question

Two persons $A$ and $B$ are located X-Y plane at the points $(0, 0)$ and $(0, 10)$ respectively. (The distances are measured in MKS units). At time $t = 0$ they start moving simultaneously with velocities $_{A} = 2^j {m s}^{- 1}$ and $_{B} = 2^i {m s}^{- 1}$ respectively. Find the time after which $A$ and $B$ are at their closest distance.

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Solution

$x = V_{A} t = 2 t \land i y = 2 t \land j$
distance between
A and B $∥$ S
$s = \sqrt{y^{2} + {(10 - x)}^{2}} = \sqrt{4 t^{2} + {(10 - 2 t)}^{2}} s^{2} = {4 t}^{2} + {4 t}^{2} - 4 t + 100 = 8 t^{2} - 4 t + 100$
For minimum $S . \frac{d s}{d t} = 0 t = \frac{4}{16} = \frac{1}{4} s e c$

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