91Ó°ÊÓ

Vector calculation can be used to find the dot product and cross product. 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.

First, find the value for each vector with their corresponding coefficient, and solve for the remaining terms using the formula for dot and cross product. The formula for the cross product is as below:

$\stackrel{→}{A}×\stackrel{→}{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ A_x& A_y& A_z\\ B_x& B_y& B_z\end{array}\right|$ (i)

The formula for the dot product is as below:

$\stackrel{→}{A}.\stackrel{→}{B}=A×B\mathrm{³\ae ³¦´Ç²õθ}$ (ii)

The given quantities are,

$$\begin{gathered} \stackrel{→}{A}=2.0\hat{i}+3.0\hat{j}-4.0\hat{k} \\ \stackrel{→}{B}=3.0\hat{i}+4.0\hat{j}+2.0\hat{k} \\ \stackrel{→}{C}=7.0\hat{i}-3.0\hat{j}+0\hat{k} \end{gathered}$$

Calculate $3\stackrel{→}{C}$and $2\stackrel{→}{A}$by using the given value of vectors.

$$\begin{gathered} 3\stackrel{→}{C}=3×\left(7.0\hat{i}-8.0\hat{j}+0\hat{k}\right) \\ =21.0\hat{i}-24.0\hat{j}+0\hat{k} \end{gathered}$$

$$\begin{gathered} 2\stackrel{→}{C}=2×\left(2.0\hat{i}+3.0\hat{j}-4.0\hat{k}\right) \\ =4.0\hat{i}+6.0\hat{j}-8.0\hat{k} \end{gathered}$$

Thus, the vector $3\stackrel{→}{C}$is $21.0\hat{i}-24.0\hat{j}+0\hat{k}$and vector$2\stackrel{→}{A}$ is $4.0\hat{i}+6.0\hat{j}-8.0\hat{k}$.

Now, calculate $\left(2\stackrel{→}{A}×\stackrel{→}{B}\right)$by substituting the value$2\stackrel{→}{A}$from step (i) and value of$\stackrel{→}{B}$from the given quantities.

$$\begin{gathered} \left(2\stackrel{→}{A}×\stackrel{→}{B}\right)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2A_x& 2A_y& 2A_z\\ B_x& B_y& B_z\end{array}\right| \\ =\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 6& -8\\ -3& 4& 2\end{array}\right| \\ =44\hat{i}+16\hat{j}+34\hat{k} \end{gathered}$$

Calculate the dot product of $3\stackrel{→}{C}$and $\left(2\stackrel{→}{A}×\stackrel{→}{B}\right)$by using the values calculated in step 2 and above.

$$\begin{gathered} 3\stackrel{→}{A}.\left(2\stackrel{→}{A}×\stackrel{→}{B}\right)=\left(21.0\hat{i}+24.0\hat{j}+0\hat{k}\right).\left(44\hat{i}+16\hat{j}+34\hat{k}\right) \\ =\left(\left(21×44\right)\right)+\left(\left(-24×16\right)\right)+\left(\left(0×34\right)\right) \\ =540\mathrm{units} \end{gathered}$$

Thus, the value of $3\stackrel{→}{A}.\left(2\stackrel{→}{A}×\stackrel{→}{B}\right)$is$540\mathrm{units}$ .