Question

$$\begin{gathered} \mathit{Q}\mathbf{}\mathbf{26}\mathbf{.}\mathbf{}Threeforceshavingmagnitudes5,4and3unitsactonaparticleinthedirections2\overline{i}-2\overline{j}-\overline{k},\overline{i}+2\overline{j}+2\overline{k}and-2\overline{i}+\overline{j}-2\overline{k}respectivelyandtheparticlegetsdisplacedfromthepointAwhosepositionvectoris6\overline{i}-2\overline{j}+3\overline{k}tothepointBwhosepositionvectoris9\overline{i}+7\overline{j}+5\overline{k}thentheworkdonebytheseforcesis \\ \left(a\right)9unit\left(b\right)43unit\left(c\right)30unit\left(d\right)45unit \end{gathered}$$

Answer 1

Dear student,
$$\begin{gathered} \overrightarrow{A_1}=2i-2j-k,\overrightarrow{B_1}=i+2j+2k,\overrightarrow{C_1}=-2i+j-2k \\ \left|\overrightarrow{A_1}\right|=\sqrt{2^2+2^2+1^2}=3=\left|\overrightarrow{B_1}\right|=\left|\overrightarrow{C_1}\right| \\ \widehat{A_1}=\frac{2i-2j-k}{3},\widehat{B_1}=\frac{i+2j+2k}{3},\widehat{C_1}=\frac{-2i+j-2k}{3} \\ thenforceshavingmagnitudes5,4and3 \\ \overrightarrow{F_1}=\frac{5\left(2i-2j-k\right)}{3},\overrightarrow{F_2}=\frac{4\left(i+2j+2k\right)}{3},\overrightarrow{F_3}=\frac{3\left(-2i+j-2k\right)}{3} \\ totalforce\overrightarrow{F}=\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}=\frac{5\left(2i-2j-k\right)}{3}+\frac{4\left(i+2j+2k\right)}{3}+\frac{3\left(-2i+j-2k\right)}{3} \\ \overrightarrow{F}=\frac{5\left(2i-2j-k\right)+4\left(i+2j+2k\right)+3\left(-2i+j-2k\right)}{3}=\frac{i\left(10+4-6\right)+j\left(-10+8+3\right)+k\left(-5+8-6\right)}{3} \\ \overrightarrow{F}=\frac{8i+j-3k}{3} \\ displacementvector \\ \overrightarrow{d}=\overrightarrow{B}-\overrightarrow{A}=9i+7j+5k-6i+2j-3k=3i+9j+2k \\ work=\overrightarrow{F}.\overrightarrow{d}=\frac{8i+j-3k}{3}.\left(3i+9j+2k\right)=8+3-2=9unitanswer \end{gathered}$$
Regards