Q180 / 180 · COMEDK 2020 · Physics

A particle starts moving from point $(2,10,1)$. Displacement for the particle is $8 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$. The final coordinates of the particle is

  • Correct answer
    A
    $(10,8,2)$
  • B
    $(8,10,2)$
  • C
    $(2,10,8)$
  • D
    $(8,2,10)$

Solution

Position vector of starting point $(2,10,1)$ of particle is given as $$ \mathbf{r}_{1}=2 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}+\hat{\mathbf{k}} $$ If final co-ordinate of the particle be $(x, y, z)$, then its position vector $$ \begin{gathered} \mathbf{r}_{2}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ \therefore \text { Displacement, } \Delta \mathbf{r}=\mathbf{r}_{2}-\mathbf{r}_{1} \\ 8 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}-(2 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ \Rightarrow \quad 8 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}=(x-2) \hat{\mathbf{i}}+(y-10) \hat{\mathbf{j}}+(z-1) \hat{\mathbf{k}} \end{gathered} $$ Comparing the coefficients of $\hat{\mathbf{i}}, \hat{\mathfrak{j}}$ and $\hat{\mathbf{k}}$, we get $$ \begin{aligned} & x-2=8, y-10=-2, z-1=1 \\ \Rightarrow & x=10, y=8, z=2 \end{aligned} $$ $\therefore$ Final co-ordinate $=(10,8,2)$

Hence, the correct answer is option: 'A'.

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