91Ó°ÊÓ

Vector operations can be used to find the dot product, cross product, and addition between two vectors. The addition of vectors gives another vector quantity. The cross product of two vectors results in a vector quantity that is perpendicular to both vectors whereas the dot product of two vectors produces a scalar quantity.

The formula for the cross product is,

$\stackrel{→}{a}×\stackrel{→}{b}=(a_y{b}_z-b_y{a}_z)\hat{i}+(a_z{b}_x-b_z{a}_x)\hat{j}+(a_x{b}_y-b_x{a}_y)\hat{k}$ (1)

The formula for the dot product is,

$\stackrel{→}{a}.\stackrel{→}{b}=a×b×\cos \mathrm{θ}$ (2)

First, find the individual values of each term like $(\stackrel{→}{\mathrm{b}}×\stackrel{→}{\mathrm{c}})$and $(\stackrel{→}{b}+\stackrel{→}{c})$. After that using the formula for the dot and cross product of vectors, get the required answer.

The vectors are given below:

$$\begin{gathered} \stackrel{→}{a}=3.0\hat{i}+3.0\hat{j}-2.0\hat{k} \\ \hat{b}=-1.0\hat{i}-4.0\hat{j}+2.0\hat{k} \\ \stackrel{→}{c}=2.0\hat{i}+2.0\hat{j}+1.0\hat{k} \end{gathered}$$

$$\begin{gathered} \mathrm{Write}\mathrm{the}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{for}\stackrel{→}{\mathrm{b}}\mathrm{and}\stackrel{→}{\mathrm{c}}. \\ (\stackrel{→}{\mathrm{b}}×\stackrel{→}{\mathrm{c}})=({\mathrm{b}}_{\mathrm{y}}{\mathrm{c}}_{\mathrm{z}}-{\mathrm{c}}_{\mathrm{y}}{\mathrm{b}}_{\mathrm{z}})\hat{\mathrm{i}}+({\mathrm{b}}_{\mathrm{z}}{\mathrm{c}}_{\mathrm{x}}-{\mathrm{b}}_{\mathrm{x}}{\mathrm{c}}_{\mathrm{z}})\hat{\mathrm{j}}+({\mathrm{b}}_{\mathrm{x}}{\mathrm{c}}_{\mathrm{y}}-{\mathrm{b}}_{\mathrm{y}}{\mathrm{c}}_{\mathrm{x}})\hat{\mathrm{k}} \\ =-8.0\hat{\mathrm{i}}+5.0\hat{\mathrm{j}}+6.0\hat{\mathrm{k}} \\ \mathrm{Find}\mathrm{the}\mathrm{dot}\mathrm{product}\mathrm{of}\stackrel{→}{\mathrm{b}}×\stackrel{→}{\mathrm{c}}\mathrm{with}\stackrel{→}{\mathrm{a}}. \\ \stackrel{→}{\mathrm{a}}.(\stackrel{→}{\mathrm{b}}×\stackrel{→}{\mathrm{c}})=\left(\right(3.0×8.0\left)\right)+\left(\right(3.0×5.0\left)\right)+\left(\right(-2.0×6.0\left)\right) \\ =-21 \\ \mathrm{Thus},\mathrm{the}\mathrm{value}\mathrm{of}\stackrel{→}{\mathrm{a}}.(\stackrel{→}{\mathrm{b}}×\stackrel{→}{\mathrm{c}})\mathrm{is}-21. \end{gathered}$$

$$\begin{gathered} \mathrm{First},\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\stackrel{→}{\mathrm{b}}+\stackrel{→}{\mathrm{c}}. \\ (\stackrel{→}{\mathrm{b}}+\stackrel{→}{\mathrm{c}})=(-1.0+2.0)\hat{\mathrm{i}}+(-4.0+2.0)\hat{\mathrm{j}}+(2.0+1.0)\hat{\mathrm{k}} \\ =1.0\hat{\mathrm{i}}+2.0\hat{\mathrm{j}}+3.0\hat{\mathrm{k}} \\ (\stackrel{→}{\mathrm{b}}+\stackrel{→}{\mathrm{c}})=1.0\hat{\mathrm{i}}+2.0\hat{\mathrm{j}}+3.0\hat{\mathrm{k}}\left(3\right) \\ \mathrm{Now},\mathrm{find}\mathrm{the}\mathrm{dot}\mathrm{product}\mathrm{of}\stackrel{→}{\mathrm{a}}\mathrm{with}\stackrel{→}{\mathrm{b}}+\stackrel{→}{\mathrm{c}}. \\ \stackrel{→}{\mathrm{a}}.(\stackrel{→}{\mathrm{b}}+\stackrel{→}{\mathrm{c}})=\left(\right(3.0×1.0\left)\right)+\left(\right(3.0×-2.0\left)\right)+\left(\right(-2.0×3.0\left)\right) \\ =-9.0 \\ \mathrm{Thus},\mathrm{the}\mathrm{value}\mathrm{of}\stackrel{→}{\mathrm{a}}.(\stackrel{→}{\mathrm{b}}+\stackrel{→}{\mathrm{c}})\mathrm{is}-9.0. \end{gathered}$$

$$\begin{gathered} \mathrm{Use}\mathrm{the}\mathrm{equation}\left(\mathrm{iii}\right)\mathrm{to}\mathrm{write}\mathrm{the}\mathrm{cross}\mathrm{product}\mathrm{of}\stackrel{→}{\mathrm{a}}\mathrm{and}\stackrel{→}{\mathrm{b}}+\stackrel{→}{\mathrm{c}}. \\ \stackrel{→}{\mathrm{a}}×(\stackrel{→}{\mathrm{b}}+\stackrel{→}{\mathrm{c}})=\left(3.0\hat{\mathrm{i}}+3.0\hat{\mathrm{j}}-2.0\hat{\mathrm{k}}\right)×\left(1.0\hat{\mathrm{i}}-2.0\hat{\mathrm{j}}+3.0\hat{\mathrm{k}}\right) \\ \mathrm{Calculate}\mathrm{the}\mathrm{cross}\mathrm{product}\mathrm{using}\mathrm{equation}\left(\mathrm{i}\right) \\ \stackrel{→}{\mathrm{a}}×(\stackrel{→}{\mathrm{b}}+\stackrel{→}{\mathrm{c}})=5.0\hat{\mathrm{i}}-11\hat{\mathrm{j}}-9.0\hat{\mathrm{k}} \\ \mathrm{Therefore},\mathrm{the}\mathrm{value}\mathrm{of}\stackrel{→}{\mathrm{a}}×(\stackrel{→}{\mathrm{b}}+\stackrel{→}{\mathrm{c}})\mathrm{is}5\hat{\mathrm{i}}-11\hat{\mathrm{j}}-9\hat{\mathrm{k}} \end{gathered}$$