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Q.

At any instant, the position and velocity vectors of two particles are $\overrightarrow {{r_1}} ,\,\overrightarrow {{r_2}} \,and\,\overrightarrow {{v_1}} ,\,\overrightarrow {{v_2}}$ respectively. They will collide if :

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a

${\bar v_1} = {\bar v_2}\$

b

$({\bar r_1} - {\bar r_2}) \cdot ({\bar v_1} - {\bar v_2}) = 0\$

c

$({\bar r_1} - {\bar r_2}) \times ({\bar v_1} - {\bar v_2}) = 0$

d

${r_1} > {r_2};{v_1} < {v_2}\$

answer is C.

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Detailed Solution

Since initial position vectors are ${\bar r_1},{\bar r_2}$ therefore position vector of the particles after time t will be $({\bar r_1} + {\bar v_1}t)\,\,and\,\,({\bar r_2} + {\bar v_2}t)$ . At the time of collision, the position vectors of both particles should be same.

${\bar r_1} + {\bar v_1}t = {\bar r_2} + {\bar v_2}t$

$({\bar r_1} - {\bar r_2}) = - ({\bar v_1} - {\bar v_2})t$

$({\bar r_1} - {\bar r_2}) \times ({\bar v_1} - {\bar v_2}) = - ({\bar v_1} - {\bar v_2}) \times ({\bar v_1} - {\bar v_2})t$

$({\bar r_1} - {\bar r_2}) \times ({\bar v_1} - {\bar v_2}) = 0$

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At any instant, the position and velocity vectors of two particles are respectively. They will collide if :