Question

The work done by the force $\overrightarrow{F}=z\hat{i}+x\hat{j}+y\hat{k}$ in taking a particle (in one second) from the origin to the point $(x_1,y_1,z_1)$ by the shortest path is:

• A. $\frac{1}{3}(x_1^2{y}_1+y_1^2{z}_1+z_1^2{x}_1)$
• B. $\frac{1}{2}(x_1{y}_1+y_1{z}_1+z_1{x}_1)$
• C. $\frac{1}{3}(x_1{y}_1+y_1{z}_1+z_1{x}_1)$
• D. $\frac{1}{4}(x_1{y}_1+y_1{z}_1-z_1{x}_1)$

Answer 1

Answer: (B)

The shortest path is along the line OP joining (0,0,0) and ($x_1,y_1,z_1$.)

The equation of the line OP is
$\frac{x-0}{0-x_1}=\frac{y-0}{0-y_1}=\frac{z-0}{0-z_1}$

$\Rightarrow \frac{x}{x_1}=\frac{y}{y_1}=\frac{z}{z_1}=t$

$dx=x_{1}dt$
$dy=y_{1}dt$
$dz=z_{1}dt$

W$=$work done by the force$F={\int }_0^{p}F.dr$

$\Rightarrow W={\int }_{(0,0,0)}^{(x_1,y_1,z_1)}(z\hat{i}+x\hat{j}+y\hat{k}).(dx\hat{i}+dy\hat{j}+dz\hat{k})$

${\int }_{(0,0,0)}^{(x_1,y_1,z_1)}zdx+xdy+ydz$

When $x=y=z=0,t=0$

When$x=x_1,y=y_1,z=z_1,t=1$

$\Rightarrow W={\int }_{t=0}^{t=1}t{z}_1{x}_{1}dt+t{x}_1{y}_{1}dt+t{y}_1{z}_{1}dt$

$={\int }_{t=0}^{t=1}(x_1{z}_1+x_1{y}_1+y_1{z}_1)tdt$

$=(x_1{y}_1+y_1{z}_1+z_1{x}_1){\int }_0^{1}tdt$

$=(x_1{y}_1+y_1{z}_1+z_1{x}_1){\left[\frac{t^2}{2}\right]}_0^1$

$=\frac{1}{2}(x_1{y}_1+y_1{z}_1+z_1{x}_1)$