Question

Given $\mathit{M}\mathbf{=}\mathbf{2}\widehat{\mathbf{i}}\mathbf{-}\mathbf{3}\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}}$and $\mathit{N}\mathbf{=}\mathbf{4}\widehat{\mathbf{i}}\mathbf{+}\mathbf{5}\widehat{\mathbf{j}}\mathbf{-}\mathbf{2}\widehat{\mathbf{k}}$calculate the vector product$\overrightarrow{\mathbf{M}}\mathbf{\times }\overrightarrow{\mathbf{N}}$.

Short answer

The vector product is $\overrightarrow{M}\times \overrightarrow{N}=(1,8,22)$.

Step-by-step solution

Define vector product:

A vector has both the direction (indicate by an arrow) and the magnitude (represented by the length).

The product of the vector is a result of combining two vectors, namely scalar and vector. As a result, you have two options for multiplying the vectors. The first is the Scalar product, which is also known as the dot product of vectors. Another is the vector product, which is also known as the cross-product of vectors.

To find the vector product:

The direction is supplied by the right-hand rule and is perpendicular to both A and B magnitude calculations.

The multiplication of the vectors, $\overrightarrow{A}\times \overrightarrow{B}$ is perpendicular to both A and B magnitude calculations. If the vectors are expressed in $\widehat{i},\widehat{j},\widehat{k}$in the $x,y,z$ directions respectively.

$\overrightarrow{A}\times \overrightarrow{B}=\left[\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ A_x& A_y& A_z\\ {\text{B}}_x& B_y& B_z\end{array}\right]$

$=\widehat{i}(A_y{B}_z-A_z{B}_y)-\widehat{j}(A_x{B}_z-A_z{B}_x)+\widehat{k}(A_x{B}_y-A_y{B}_x)$

Therefore by using the above expression, you can define $\overrightarrow{M}\times \overrightarrow{N}$ as follow.

$\begin{array}{rcl}\overrightarrow{M}\times \overrightarrow{N}& =& \left(\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 2& -3& 1\\ 4& 5& -2\end{array}\right)\\ & & \begin{array}{c}=\widehat{i}[((-3)\times (-2))-(1\times 5)]-\widehat{j}[((-2)\times 2)-(4\times 1)]+\widehat{k}[(2\times 5)-((-3)\times 4)]\\ =\widehat{i}[6-5]-\widehat{j}[-4-4]+\widehat{k}[10+12]\\ =\widehat{i}+8\widehat{j}+22\widehat{k}\end{array}\end{array}$

Hence, the vector product is$\overrightarrow{M}\times \overrightarrow{N}=(1,8,22)$