Question

Grade 12 vectors Part 1 Q2Items a), b) and c)

Answer 1

By definition, the vectorial product of two vectors with the following coordinates:

$\begin{gathered} \vec{u}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k} \\ \vec{v}=x_2\hat{i}+y_2\hat{j}+z_2\hat{k} \end{gathered}$

The vectorial product between those two vectors is given by the following determinant:

$\begin{gathered} \vec{u}\times\vec{v}=\det\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {x_1} & {y_1} & {z_1} \\ {x_2} & {y_2} & {z_2}\end{bmatrix} \\ =(y_1z_2-y_2z_1)\hat{i}+(x_2z_1-x_1z_2)\hat{j}+(x_1y_2-x_2y_1)\hat{k} \end{gathered}$

item a):

Using the previous definition in our problem, we have:

$\vec{k}\times\vec{j}=\det\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {0} & {0} & {1} \\ {0} & {1} & {0}\end{bmatrix}=(0-1)\hat{i}+(0-0)\hat{j}+(0-0)\hat{k}=-\hat{i}$

item b):

$(\hat{i}\times\hat{j})\times\hat{i}=\det\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {1} & {0} & {0} \\ {0} & {1} & {0}\end{bmatrix}\times\hat{i}=\hat{k}\times\hat{i=\hat{j}}$

item c):

The vectorial product of a vector and itself is always equal to zero.

$\vec{a}\times\vec{a}=0$