Question

Two vectors are given by,$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{)}\hat{\mathbf{i}}\mathbf{-}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{)}\hat{\mathbf{j}}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{)}\hat{\mathbf{k}}$And$\stackrel{\mathbf{\rightharpoonup }}{\mathit{b}}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{)}\hat{\mathbf{j}}\mathbf{+}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{)}\hat{\mathbf{k}}$In unit-vector notation, findrole="math" localid="1656159321000" $\left(a\right)\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}\mathbf{+}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{b}}\mathbf{,}\left(b\right)\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}\mathbf{-}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{b}}$and a third vector$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{c}}$such that$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}\mathbf{-}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{b}}\mathbf{=}\mathbf{0}$

Short answer

$$\begin{gathered} \mathit{a}\mathbf{)}\mathbf{}\mathbf{}\stackrel{\rightharpoonup }{\mathrm{a}}+\stackrel{\rightharpoonup }{\mathrm{b}},=\left(3\right)\stackrel{\rightharpoonup }{\mathrm{i}}-\left(2\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(5\right)\stackrel{\rightharpoonup }{\mathrm{k}} \\ \mathbf{b}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\stackrel{\rightharpoonup }{\mathrm{a}}+\stackrel{\rightharpoonup }{\mathrm{b}},=\left(5\right)\stackrel{\rightharpoonup }{\mathrm{i}}-\left(4\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(3\right)\stackrel{\rightharpoonup }{\mathrm{k}} \\ \mathbf{c}\mathbf{)}\stackrel{\rightharpoonup }{\mathrm{c}}=\left(-5\right)\stackrel{\rightharpoonup }{\mathrm{i}}+\left(4\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(3\right)\stackrel{\rightharpoonup }{\mathrm{k}} \end{gathered}$$

Step-by-step solution

To understand the concept of vector addition and subtraction

Components of all the vectors which are associated with the same unit can be added or subtracted. The addition and subtraction of the vectors can be written as

$$\begin{gathered} \stackrel{\rightharpoonup }{\mathrm{a}}+\stackrel{\rightharpoonup }{\mathrm{b}}=\left(A_x+B_x\right)\stackrel{\rightharpoonup }{\mathrm{I}}+\left(A_y+B_y\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(A_z+B_z\right)\stackrel{\rightharpoonup }{\mathrm{k}} \\ \stackrel{\rightharpoonup }{\mathrm{a}}+\stackrel{\rightharpoonup }{\mathrm{b}}=\left(A_x-B_x\right)\stackrel{\rightharpoonup }{\mathrm{I}}+\left(A_y-B_y\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(A_z-B_z\right)\stackrel{\rightharpoonup }{\mathrm{k}} \end{gathered}$$

Given

$$\begin{gathered} \stackrel{\rightharpoonup }{a}=\left(4\right)\stackrel{\rightharpoonup }{\mathrm{i}}-\left(3\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(1\right)\stackrel{\rightharpoonup }{\mathrm{k}} \\ \stackrel{\rightharpoonup }{b}=\left(-1\right)\stackrel{\rightharpoonup }{\mathrm{i}}-\left(1\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(4\right)\stackrel{\rightharpoonup }{\mathrm{k}} \end{gathered}$$

To finda⇀+b⇀  

$\stackrel{\rightharpoonup }{a}+\stackrel{\rightharpoonup }{b}=\left[\left(4\right)\stackrel{\rightharpoonup }{\mathrm{i}}-\left(3\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(1\right)\stackrel{\rightharpoonup }{\mathrm{k}}\right]+\left[\left(-1\right)\stackrel{\rightharpoonup }{\mathrm{i}}-\left(1\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(4\right)\stackrel{\rightharpoonup }{\mathrm{k}}\right]$

According to vector algebra, we can add or subtract vectors in same direction. Thus,

$\stackrel{\rightharpoonup }{a}+\stackrel{\rightharpoonup }{b}=\left(3\right)\stackrel{\rightharpoonup }{\mathrm{i}}-\left(2\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(5\right)\stackrel{\rightharpoonup }{\mathrm{k}}$

To finda⇀-b⇀  

$\stackrel{\rightharpoonup }{a}-\stackrel{\rightharpoonup }{b}=\left[\left(4\right)\stackrel{\rightharpoonup }{\mathrm{i}}-\left(3\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(1\right)\stackrel{\rightharpoonup }{\mathrm{k}}\right]+\left[\left(-1\right)\stackrel{\rightharpoonup }{\mathrm{i}}-\left(1\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(4\right)\stackrel{\rightharpoonup }{\mathrm{k}}\right]$

Thus

$\stackrel{\rightharpoonup }{a}-\stackrel{\rightharpoonup }{b}=\left(5\right)\stackrel{\rightharpoonup }{\mathrm{i}}-\left(4\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(3\right)\stackrel{\rightharpoonup }{\mathrm{k}}$

To findc⇀  

$$\begin{gathered} \stackrel{\rightharpoonup }{a}-\stackrel{\rightharpoonup }{b}+\stackrel{\rightharpoonup }{c}=0 \\ \stackrel{\rightharpoonup }{c}=\stackrel{\rightharpoonup }{b}-\stackrel{\rightharpoonup }{a} \\ \stackrel{\rightharpoonup }{c}=\left(-1\right)\stackrel{\rightharpoonup }{\mathrm{i}}-\left(1\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(4\right)\stackrel{\rightharpoonup }{\mathrm{k}}-\left(\left(-4\right)\stackrel{\rightharpoonup }{\mathrm{i}}-\left(3\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(1\right)\stackrel{\rightharpoonup }{\mathrm{k}}\right) \\ \stackrel{\rightharpoonup }{c}=\left(-5\right)\stackrel{\rightharpoonup }{\mathrm{i}}+\left(4\right)\stackrel{\rightharpoonup }{\mathrm{j}}+\left(3\right)\stackrel{\rightharpoonup }{\mathrm{k}} \end{gathered}$$