Question

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

(i + j) × (i - j)

and

k × (i - 2j)

Answer 1

$(\mathbf i+\mathbf j)\times(\mathbf i-\mathbf j)=(\mathbf i+\mathbf j)\times\mathbf i+(\mathbf i+\mathbf j)\times(-\mathbf j)$

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

$(\mathbf i+\mathbf j)\times\mathbf i+(\mathbf i+\mathbf j)\times(-\mathbf j)=-(\mathbf i\times(\mathbf i+\mathbf j))-(-\mathbf j)\times(\mathbf i+\mathbf j))$

Scalars can be factored, so

$-(\mathbf i\times(\mathbf i+\mathbf j))-(-\mathbf j)\times(\mathbf i+\mathbf j))=-(\mathbf i\times(\mathbf i+\mathbf j))+\mathbf j\times(\mathbf i+\mathbf j))$

Using the distributive property again,

$-(\mathbf i\times(\mathbf i+\mathbf j))+\mathbf j\times(\mathbf i+\mathbf j))=-(\mathbf i\times\mathbf i+\mathbf i\times\mathbf j)+\mathbf j\times\mathbf i+\mathbf j\times\mathbf j$

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

$-(\mathbf i\times\mathbf i+\mathbf i\times\mathbf j)+\mathbf j\times\mathbf i+\mathbf j\times\mathbf j=-(\mathbf i\times\mathbf j)+\mathbf j\times\mathbf i$

Because $\mathbf i\times\mathbf j=\mathbf k$, and by anticommutativity, you are left with

$-(\mathbf i\times\mathbf j)+\mathbf j\times\mathbf i=-\mathbf k-(\mathbf i\times\mathbf j)=-\mathbf k-\mathbf k=-2\mathbf k$

Next, without outlining which properties are being used,

$\mathbf k\times(\mathbf i-2\mathbf j)=\mathbf k\times\mathbf i+2(\mathbf j\times\mathbf k)=2\mathbf i+\mathbf j$