The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

\(\displaystyle \widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}\)

\(\displaystyle \widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}\)

\(\displaystyle \widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}\)

\(\displaystyle \widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0\)

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

\(\displaystyle r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}\)

With these principles in mind, we can calculate the cross product of our vectors \(\displaystyle \overrightarrow{a}=4\widehat{i}-4\widehat{j}+3\widehat{k}\) and \(\displaystyle \overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k}\)

\(\displaystyle \overrightarrow{a}\times \overrightarrow{b}=(-12-6)\widehat{i}+(-12+3)\widehat{j}+(8+4)\widehat{k}=-18\widehat{i}-9\widehat{j}+12\widehat{k}\)

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

\(\displaystyle \widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}\)

\(\displaystyle \widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}\)

\(\displaystyle \widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}\)

\(\displaystyle \widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0\)

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

\(\displaystyle r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}\)

With these principles in mind, we can calculate the cross product of our vectors \(\displaystyle \overrightarrow{a}=20\widehat{i}+2\widehat{j}\) and \(\displaystyle \overrightarrow{b}=15\widehat{i}-8\widehat{j}\)

\(\displaystyle \overrightarrow{a}\times \overrightarrow{b}=(0-160-30+0)\widehat{k}=-190\widehat{k}\)

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

\(\displaystyle \widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}\)

\(\displaystyle \widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}\)

\(\displaystyle \widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}\)

\(\displaystyle \widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0\)

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

\(\displaystyle r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}\)

With these principles in mind, we can calculate the cross product of our vectors \(\displaystyle \overrightarrow{a}=6\widehat{i}-\widehat{j}\) and \(\displaystyle \overrightarrow{b}=11\widehat{i}+5\widehat{j}\)

\(\displaystyle \overrightarrow{a}\times \overrightarrow{b}=(0+30+11+0)\widehat{k}=41\widehat{k}\)

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

\(\displaystyle \widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}\)

\(\displaystyle \widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}\)

\(\displaystyle \widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}\)

\(\displaystyle \widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0\)

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

\(\displaystyle r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}\)

With these principles in mind, we can calculate the cross product of our vectors \(\displaystyle \overrightarrow{a}=9\widehat{i}-9\widehat{j}\) and \(\displaystyle \overrightarrow{b}=-2\widehat{i}+2\widehat{j}\)

\(\displaystyle \overrightarrow{a}\times \overrightarrow{b}=(0+18-18+0)\widehat{k}=0\)

This zero result stems from the fact that these vectors are parallel, a fact which might be apparent from quick observation.

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

\(\displaystyle \widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}\)

\(\displaystyle \widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}\)

\(\displaystyle \widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}\)

\(\displaystyle \widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0\)

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

\(\displaystyle r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}\)

With these principles in mind, we can calculate the cross product of our vectors \(\displaystyle \overrightarrow{a}=2\widehat{i}+6\widehat{j}+3\widehat{k}\) and \(\displaystyle \overrightarrow{b}=\widehat{i}-\widehat{j}-2\widehat{k}\)

\(\displaystyle \overrightarrow{a}\times \overrightarrow{b}=(-12+3)\widehat{i}+(4+3)\widehat{j}+(-2-6)\widehat{k}=-9\widehat{i}+7\widehat{j}-8\widehat{k}\)

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

\(\displaystyle \widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}\)

\(\displaystyle \widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}\)

\(\displaystyle \widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}\)

\(\displaystyle \widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0\)

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

\(\displaystyle r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}\)

With these principles in mind, we can calculate the cross product of our vectors \(\displaystyle \overrightarrow{a}=14\widehat{i}+5\widehat{j}\) and \(\displaystyle \overrightarrow{b}=4\widehat{i}+3\widehat{j}\)

\(\displaystyle \overrightarrow{a}\times \overrightarrow{b}=(0+42-20+0)\widehat{k}=22\widehat{k}\)

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

\(\displaystyle \widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}\)

\(\displaystyle \widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}\)

\(\displaystyle \widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}\)

\(\displaystyle \widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0\)

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

\(\displaystyle r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}\)

With these principles in mind, we can calculate the cross product of our vectors \(\displaystyle \overrightarrow{a}=9\widehat{j}\) and \(\displaystyle \overrightarrow{b}=3\widehat{j}\)

\(\displaystyle \overrightarrow{a}\times \overrightarrow{b}=(27(0))\widehat{k}=0\)

It may go without saying that these two vectors are parallel (afterall, both go strictly in the j-direction), and so the cross product is zero.

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

\(\displaystyle \widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}\)

\(\displaystyle \widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}\)

\(\displaystyle \widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}\)

\(\displaystyle \widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0\)

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

\(\displaystyle r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}\)

With these principles in mind, we can calculate the cross product of our vectors \(\displaystyle \overrightarrow{a}=2\widehat{i}-6\widehat{j}\) and \(\displaystyle \overrightarrow{b}=\widehat{i}-7\widehat{j}\)

\(\displaystyle \overrightarrow{a}\times \overrightarrow{b}=(0-14+6+0)\widehat{k}=-8\widehat{k}\)

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

\(\displaystyle \widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}\)

\(\displaystyle \widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}\)

\(\displaystyle \widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}\)

\(\displaystyle \widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0\)

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

\(\displaystyle r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}\)

With these principles in mind, we can calculate the cross product of our vectors \(\displaystyle \overrightarrow{a}=2\widehat{i}+\widehat{j}\) and \(\displaystyle \overrightarrow{b}=\widehat{i}+3\widehat{j}\)

\(\displaystyle \overrightarrow{a}\times \overrightarrow{b}=(6-1)\widehat{k}=5\widehat{k}\)

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

\(\displaystyle \widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}\)

\(\displaystyle \widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}\)

\(\displaystyle \widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}\)

\(\displaystyle \widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0\)

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

\(\displaystyle r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}\)

With these principles in mind, we can calculate the cross product of our vectors \(\displaystyle \overrightarrow{a}=2\widehat{j}\) and \(\displaystyle \overrightarrow{b}=3\widehat{i}+4\widehat{j}\)

\(\displaystyle \overrightarrow{a}\times \overrightarrow{b}=(-6+0)\widehat{k}=-6\widehat{k}\)