# Find the vector, not with determinants, but by using properties of cross products. (i+j)\times(i-j)

Vectors
Find the vector, not with determinants, but by using properties of cross products. $$(i+j)\times(i-j)$$

2021-06-10
Calculations:
Apply the distribution property of the cross product, we get
$$(i+j)\times(i-j)=i\times i-i\times j+j\times i-j\times j$$
And since the angle between the cross product $$i\times i$$ and $$j\times j$$ is zero, and since $$\sin(0^\circ)=0$$ then from equation the cross product for any two vectors having the same direction is also zero, substituting we get
$$=0-i\times j+j\times i-0$$
$$=-i\times j+j\times i$$
Applying, the commutation property of the cross product, we get
$$=-i\times j-i\times j$$
Applying equation, we get
$$=-2(|a||b|\sin(\theta)n)$$
And since the magnitude of the unit direction i and j is 1, and the angle between both vectors is $$90^\circ$$, thus $$\sin\theta=1$$ therefor we have
$$=-2(1)(1)(1)n$$
$$i\times j=-2n$$
And, the direction of the unit vector n is orthogonal to $$i\times j$$, is determined using the right hand rule, where curling the fingers from the i to j the thumb would be pointing toward the positive direction of k, therefor
$$n=k$$
And, thus the final answer is
$$=-2k$$
Result: -2k