Question

Find the vectors, not with determinants, but by using properties of cross products.

(i + j) × (i - j)

and

k × (i - 2j)

Answer 1

`(\mathbf i+\mathbf j)*(\mathbf i-\mathbf j)=(\mathbf i+\mathbf j)*\mathbf i+(\mathbf i+\mathbf j)*(-\mathbf j)`

since the cross product is distributive over addition. By anticommutativity,


`(\mathbf i+\mathbf j)*\mathbf i+(\mathbf i+\mathbf j)*(-\mathbf j)=-(\mathbf i*(\mathbf i+\mathbf j))-(-\mathbf j)*(\mathbf i+\mathbf j))`

Scalars can be factored, so


`-(\mathbf i*(\mathbf i+\mathbf j))-(-\mathbf j)*(\mathbf i+\mathbf j))=-(\mathbf i*(\mathbf i+\mathbf j))+\mathbf j*(\mathbf i+\mathbf j))`

Using the distributive property again,


`-(\mathbf i*(\mathbf i+\mathbf j))+\mathbf j*(\mathbf i+\mathbf j))=-(\mathbf i*\mathbf i+\mathbf i*\mathbf j)+\mathbf j*\mathbf i+\mathbf j*\mathbf j`

Any vector crossed with itself is the zero vector, so you get


`-(\mathbf i*\mathbf i+\mathbf i*\mathbf j)+\mathbf j*\mathbf i+\mathbf j*\mathbf j=-(\mathbf i*\mathbf j)+\mathbf j*\mathbf i`

Because
`\mathbf i*\mathbf j=\mathbf k`, and by anticommutativity, you are left with


`-(\mathbf i*\mathbf j)+\mathbf j*\mathbf i=-\mathbf k-(\mathbf i*\mathbf j)=-\mathbf k-\mathbf k=-2\mathbf k`

Next, without outlining which properties are being used,


`\mathbf k*(\mathbf i-2\mathbf j)=\mathbf k*\mathbf i+2(\mathbf j*\mathbf k)=2\mathbf i+\mathbf j`