doing a problem about distance with vectors appears this identity:
$$\vec{A}\times\vec{B}=\frac{\vec{A}(\vec{A}\cdot\vec{B})-\vec{A}^{2}\vec{B}}{\lvert\vec{A}\lvert}=(\vec{A}\cdot\vec{B})\hat{A}-\lvert\vec{A}\lvert\vec{B}$$ I evaluate this identity in a formula that we know,and the result is ok. The formula is this:
$$\lvert\vec{A}\times\vec{B}\lvert=\lvert\vec{A}\lvert\lvert\vec{B}\lvert\sin\theta$$ we know the norm of a vector is equal to the square root of dot product between them, then we have: $$\lvert\vec{A}\times\vec{B}\lvert=\sqrt{\left[(\vec{A}\cdot\vec{B})\hat{A}-\lvert\vec{A}\lvert\vec{B}\right]\cdot\left[(\vec{A}\cdot\vec{B})\hat{A}-\lvert\vec{A}\lvert\vec{B}\right]}$$ $$=\sqrt{(\vec{A}\cdot\vec{B})^{2}-2(\vec{A}\cdot\vec{B})\lvert\vec{A}\lvert\vec{B}\cdot\hat{A}+\lvert\vec{A}\vert^{2}\lvert\vec{B}\vert^{2}}$$ $$\sqrt{-\lvert\vec{A}\vert^{2}\lvert\vec{B}\vert^{2}\cos^{2}\theta+\lvert\vec{A}\vert^{2}\lvert\vec{B}\vert^{2}}=\lvert\vec{A}\lvert\lvert\vec{B}\lvert\sqrt{1-\cos^{2}\theta}=\lvert\vec{A}\lvert\lvert\vec{B}\lvert\sin\theta$$ all look like the identity is right, Now my question is, this identity some one invent this before. Thanks in advance friends.