Let $\vec{a} , \vec{b} , \vec{c} $ three nonzero, non parallel vectors in $\mathbb{R}^3 $ for which $ (\vec{a} \times \vec{b} ) \times \vec{c} =\vec{0} $ . Prove that $\vec{a}\cdot \vec{c} = \vec{b}\cdot \vec{c} $ .

My attempt: When writing $\vec{a}=(a_1 ,a_2 , a_3 ) $, etc... , and calculating the vector product, I get that the following system must hold: $ b_1 (c_3 a_3 +c_2 a_2 ) = a_1 (c_2 b_2 +c_3 b_3 ) $

$ b_2 (c_3 a_3 +c_1 a_1 ) = a_2 (c_3 b_3 +c_1 b_1 ) $

$ b_3 (c_1 a_1 +c_2 a_2 ) = a_3 (c_1 b_1 +c_2 b_2 ) $

I know that after multiplying the first equality by $c_1$ , the second one by $c_2 $ , the third by $c_3 $ , and summing them all up , I get the same left and right hand sides, but I have no idea about what it gives me...

Will you please help me ?

Thanks in advance

  • $\begingroup$ Just a comment: In general one has the often useful identity $$a \times (b \times c) = (a \cdot c)b - (a \cdot b)c.$$ $\endgroup$ – fuglede Apr 16 '14 at 14:14

Since $ (\vec{a} \times \vec{b} ) \times \vec{c} =\vec{0} $ we must have $ (\vec{a} \times \vec{b} ) $ is parallel to $\vec c$. Also $\vec a \times \vec b$ is a vector that is perpendicular to both $\vec a $ and $\vec b$. Now since $\vec c$ is parallel to $ (\vec{a} \times \vec{b} )$ ,$\vec c$ is perpendicular to $\vec a $ and $\vec b$.

It follows that $\vec a . \vec c$=0 and $\vec b . \vec c$=0

  • $\begingroup$ Thanks a lot ! Is there any algebraic way to see this ? (i.e- without such a geometric argument) $\endgroup$ – homogenity Apr 16 '14 at 14:15
  • $\begingroup$ I am sorry but I do not see a way out of this without using the fact that $\vec a \times \vec b$ is parallel to $\vec c$ $\endgroup$ – GTX OC Apr 16 '14 at 14:23

Let's take $\vec{a} \colon= a_1 \hat{i}$, $\ \vec{b} \colon = b_1 \hat{i} + b_2 \hat{j}$, $\ \vec{c} \colon= c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k}$.

So $$(\vec{a} \times \vec{b} ) \times \vec{c} = a_1 b_2 \hat{k} \times ( c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k} ) = a_1 b_2 c_1 \hat{j} - a_1 b_2 c_2 \hat{i}.$$

Thus, if $$(\vec{a} \times \vec{b} ) \times \vec{c} = \vec{0},$$ then we must have $$a_1 b_2 c_1 \hat{j} - a_1 b_2 c_2 \hat{i} = \vec{0}.$$ And since the unit vectors $\hat{i}$ and $\hat{j}$ are non-collinear, we must have $$a_1 b_2 c_1 = 0$$ and $$a_1 b_2 c_2 = 0.$$

Now since the vector $\vec{a} \ne 0$, we must have $a_1 \ne 0$ and therefore $$ b_2 c_2 = 0$$ and $$b_2 c_1 = 0. $$

Since $\vec{a}$, $\vec{b}$, and $\vec{c}$ are non-zero, non-parallel vectors, we must have $a_1 \ne 0$, $\ b_2 \ne 0$. So we end up with $$ c_1 = c_2 = 0.$$ So $$\vec{c} = c_3 \hat{k}, $$ and $c_3 \ne 0$ since $\vec{c} \ne 0$.

Now $$\vec{a} \cdot \vec{c} = (a_1 \hat{i}) \cdot (c_3 \hat{k}) = 0.$$ And $$\vec{b} \cdot \vec{c} = (b_1 \hat{i} + b_2 \hat{j} ) \cdot (c_3 \hat{k}) = 0.$$ Hence $$\vec{a} \cdot \vec{c} = \vec{b} \cdot \vec{c},$$ as required.


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