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Question:
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A particle is acted upon by following forces:
(i) $\quad 2 \hat{i}+3 \hat{j}+5 \hat{k}$,
(ii) $-5 \hat{i}+4 \hat{j}-3 \hat{k}$ and (iii) $3 \hat{i}-7 \hat{k}$
In which plane does it move?
Options:
(i) $\quad 2 \hat{i}+3 \hat{j}+5 \hat{k}$,
(ii) $-5 \hat{i}+4 \hat{j}-3 \hat{k}$ and (iii) $3 \hat{i}-7 \hat{k}$
In which plane does it move?
Solution:
1530 Upvotes
Verified Answer
The correct answer is:
$\mathrm{y}$ -z plane
Three forces are given by, say, $\mathrm{F}_{1}, \mathrm{~F}_{2}$ and $\mathrm{F}_{3}$
$\overrightarrow{\mathrm{F}}_{1}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}, \overrightarrow{\mathrm{F}}_{2}=-5 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$
and $\vec{F}_{3}=3 \hat{i}-7 \hat{k}$
Total resultant force, $\overrightarrow{\mathrm{F}}$ is given by
$\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_{1}+\overrightarrow{\mathrm{F}}_{2}+\overrightarrow{\mathrm{F}}_{3}$
$=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}-5 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}+3 \hat{\mathrm{i}}-7 \hat{\mathrm{k}}=7 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}$
This show that the resultant force is in the $\mathrm{y}-\mathrm{z}$ plane. Thus, it moves in the $\mathrm{y}-\mathrm{z}$ plane.
$\overrightarrow{\mathrm{F}}_{1}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}, \overrightarrow{\mathrm{F}}_{2}=-5 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$
and $\vec{F}_{3}=3 \hat{i}-7 \hat{k}$
Total resultant force, $\overrightarrow{\mathrm{F}}$ is given by
$\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_{1}+\overrightarrow{\mathrm{F}}_{2}+\overrightarrow{\mathrm{F}}_{3}$
$=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}-5 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}+3 \hat{\mathrm{i}}-7 \hat{\mathrm{k}}=7 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}$
This show that the resultant force is in the $\mathrm{y}-\mathrm{z}$ plane. Thus, it moves in the $\mathrm{y}-\mathrm{z}$ plane.
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