91Ó°ÊÓ

The given vectors are $\stackrel{→}{A}=−2.00\widehat{i}+3.00\widehat{j}+4.00\widehat{k}$ and $\stackrel{→}{B}=3.00\widehat{i}+1.00\widehat{j}−3.00\widehat{k}$.

The magnitude of a vector$\stackrel{→}{V}=V_x\widehat{i}+V_y\widehat{j}+V_z\widehat{k}$is given by,

$\left|\stackrel{→}{V}\right|=\sqrt{{V_x}^2+{V_y}^2+{V_z}^2}$

Part(a)

The magnitude of vector $\stackrel{→}{A}$can be expressed as,

$\left|\stackrel{→}{A}\right|=\sqrt{A_{{}_x}^2+A_{{}_y}^2+A_{{}_z}^2}$

Substitute -2.00 for A_x ,3.00 for A_y and 4.00 for A_z

$\begin{array}{c}\stackrel{→}{A}=−2.00\widehat{i}+3.00\widehat{j}+4.00\widehat{k}\\ \left|\stackrel{→}{A}\right|=\sqrt{{\left(−2.00\right)}^2+{\left(3.00\right)}^2+{\left(4.00\right)}^2}\\ =\sqrt{29}\\ =5.38\end{array}$

The magnitude of vector$\stackrel{→}{B}$can be expressed as,

$\left|\stackrel{→}{B}\right|=\sqrt{{B_x}^2+{B_y}^2+{B_z}^2}$

Substitute 3.00 for B_x , 1.00 for B_y and -3.00 for B_z

$\begin{array}{c}\left|\stackrel{→}{B}\right|=\sqrt{{\left(3.00\right)}^2+{\left(1.00\right)}^2+{\left(−3.00\right)}^2}\\ =\sqrt{19}\\ =4.36\end{array}$

Thus, the magnitude of $\stackrel{→}{A}$ is 5.38 and magnitude of $\stackrel{→}{B}$ is 4.36.

Part(b)

The vector$\stackrel{→}{A}$is given as,

$\stackrel{→}{A}=−2.00\widehat{i}+3.00\widehat{j}+4.00\widehat{k}$and,

The vector$\stackrel{→}{B}$is given as,

$\stackrel{→}{B}=3.00\widehat{i}+1.00\widehat{j}−3.00\widehat{k}$

The difference $\stackrel{→}{A}−\stackrel{→}{B}$can be evaluated as,

$\begin{array}{c}\stackrel{→}{A}−\stackrel{→}{B}=\left(−2.00\widehat{i}+3.00\widehat{j}+4.00\widehat{k}\right)−\left(3.00\widehat{i}+1.00\widehat{j}−3.00\widehat{k}\right)\\ =\left(−2.00−3.00\right)\widehat{i}+\left(3.00−1.00\right)\widehat{j}−\left(4.00−\left(−3.00\right)\right)\widehat{k}\\ =−5.00\widehat{i}+2.00\widehat{j}+7.00\widehat{k}\end{array}$

Thus, the vector difference $\begin{array}{c}\stackrel{→}{A}−\stackrel{→}{B}\end{array}$ is $−5.00\widehat{i}+2.00\widehat{j}+7.00\widehat{k}$.

Part(c)

Consider$\stackrel{→}{A}−\stackrel{→}{B}=\stackrel{→}{G}$. The vector$\stackrel{→}{G}$can be expressed as,

$\stackrel{→}{G}=−5.00\widehat{i}+2.00\widehat{j}+7.00\widehat{k}$

Hence, the magnitude of vector$\stackrel{→}{G}$can be expressed as,

$\left|\stackrel{→}{G}\right|=\sqrt{{G_x}^2+{G_y}^2+{G_z}^2}$

Substitute -5.00 for G_x , 2.00 for G_y and 7.00 for G_z.

$\begin{array}{c}\left|\stackrel{→}{G}\right|=\sqrt{{\left(−5.00\right)}^2+{\left(2.00\right)}^2+{\left(7.00\right)}^2}\\ =\sqrt{78}\\ =8.83\end{array}$

The magnitude of$\stackrel{→}{A}−\stackrel{→}{B}$is 8.83.

The vector difference $\stackrel{→}{B}−\stackrel{→}{A}$can be evaluated as,

$\begin{array}{c}\stackrel{→}{B}−\stackrel{→}{A}=\left(3.00\widehat{i}+1.00\widehat{j}−3.00\widehat{k}\right)−\left(−2.00\widehat{i}+3.00\widehat{j}+4.00\widehat{k}\right)−\\ =\left(−3.00−\left(−2.00\right)\right)\widehat{i}+\left(1.00−3.00\right)\widehat{j}−\left(−3.00−4.00\right)\widehat{k}\\ =5.00\widehat{i}−2.00\widehat{j}−7.00\widehat{k}\end{array}$

Thus, the magnitude of$\stackrel{→}{B}−\stackrel{→}{A}$is calculated as,

$\begin{array}{c}\left|\stackrel{→}{B}−\stackrel{→}{A}\right|=\sqrt{{\left(5.00\right)}^2+{\left(−2.00\right)}^2+{\left(−7.00\right)}^2}\\ =\sqrt{78}\\ =8.83\end{array}$

Thus, $\stackrel{→}{A}−\stackrel{→}{B}$ and $\stackrel{→}{B}−\stackrel{→}{A}$ have the same magnitude.