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At a given instant of time two particles are having the position vectors $4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+57 \hat{\mathbf{k}}$ metres and $2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ respectively. If the velocity of the first particle be $0.4 \hat{\mathbf{i}} \mathrm{ms}^{-1}$, the velocity of second particle in metre per second if they collide after $10 \mathrm{sec}$ is
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The correct answer is:
$0.6\left(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right)$
$\begin{aligned} s_1+\left(u_1\right) t & =s_2+\left(u_2\right) t \\ & =(4 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}})+(0.4 \hat{\mathbf{i}}) 10 \\ & =(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})+\left(u_2\right) 10 \\ \Rightarrow \quad u_2 & =\frac{(6 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})}{10} \\ & =0.6\left[\hat{\mathbf{i}}-\hat{\mathbf{j}}+\frac{1}{3} \hat{\mathbf{k}}\right]\end{aligned}$
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