# Proof with vector properties $a = \frac{\vec{D}·(\vec{B} \times \vec{C} ) }{\vec{A} · (\vec{B} \times \vec{C} )}$

I need to prove that given 3 non parallel vectors such that $$\vec{D} = a \vec{A} + b \vec{B} + c\vec{C}$$.

We can obtain $$a = \frac{\vec{D}·(\vec{B} \times \vec{C} ) }{\vec{A} · (\vec{B} \times \vec{C} )}$$

I don't even know where to start and hope you can give me some help. Thanks.

• Hint: $\vec{B} \times \vec{C}$ is orthogonal to both $\vec{B}$ and $\vec{C}$. So what do you obtain when you taken a dot product of both the sides with $\vec{B} \times \vec{C}$? – sudeep5221 May 17 '20 at 22:28
• It's not enough that they're non-parallel, they must be linearly independent. @sudeep5221 That hint is good enough to post as an answer. – runway44 May 17 '20 at 22:31
• We get 0 since both the sides are orthogonal? Your comment made me think that by obtaining the values of a, b, c we take them to the same plane. Maybe? – Rolando González May 17 '20 at 22:35
• What equation do you get? $0$ is not an equation. And you don't get $0=0$ either. It doesn't make sense to say "both sides are orthogonal." We know $B\times C$ is orthogonal to $B$ and $C$, but we cannot say it is orthogonal to $D$ or $A$ so there are terms that remain in the equation... – runway44 May 17 '20 at 22:38

Using invariance of the triple product under circular shift one may see that $$\vec{B}\cdot[\vec{B}\times\vec{C}] = \vec{C}\cdot[\vec{B}\times\vec{B}] = 0 \\ \vec{C}\cdot[\vec{B}\times\vec{C}] = \vec{B}\cdot[\vec{C}\times\vec{C}] = 0.$$ Thus
$$\vec{D}\cdot[\vec{B}\times\vec{C}] = (a\vec{A} + b\vec{B} + c\vec{C})\cdot[\vec{B}\times\vec{C}] = a\vec{A}\cdot[\vec{B}\times\vec{C}],$$ which leads to the result required.
• I think the fact that $B\times C$ is perpendicular to $B$ and $C$ is more well-known than the circular shift invariance, honestly. – runway44 May 17 '20 at 22:37