Question 10: Motion in 2D & Vectors

Question

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

Accepted Answer

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
$$\mathbf{r}_2=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$$
$$	herefore$$ Displacement, $$\Delta \mathbf{r}=\mathbf{r}_2-\mathbf{r}_1$$
$$\begin{aligned}
& 8 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}-(2 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \
\Rightarrow \quad & 8 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}=(x-2) \hat{\mathbf{i}}+(y-10) \hat{\mathbf{j}}+(z-1) \hat{\mathbf{k}}
\end{aligned}$$
Comparing the coefficients of $$\hat{\mathbf{i}}, \hat{\mathbf{j}}$$ and $$\hat{\mathbf{k}}$$, we get
$$\begin{aligned}\quad x-2=8, y-10 & =-2, z-1=1 \
\Rightarrow \quad x=10, y & =8, z=2 \
	herefore 	ext { Final co-ordinate } & =(10,8,2)
\end{aligned}$$

Answer

(10, 8, 2)

Answer

(8, 10, 2)

Answer

(2, 10, 8)

Answer

(8, 2, 10)

Motion in 2D & Vectors

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

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