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The following problem is from the 7th edition of the book "Calculus and Analytic Geometry Part II". It can be found in section 13.7. It is problem number 12.

Problem:

Suppose that: \begin{align*} A \cdot A = 4, \,\,\,\, B \cdot B = 4, \,\,\,\,\, A \cdot B = 0 \\ (A \times B) \times C = 0, \,\,\,\,\, (A \times B) \cdot C = 8 \\ \end{align*} a) Find $A \cdot C$
(Hint: Picture the vectors, and think geometrically. Use base, coordinate-free definitions. Avoid long calculations.)

Answer:

Let $A = (a_1,a_2,a_3)$, $B = (b_1,b_2,b_3)$ and $C = (c_1,c_2,c_3)$. We have: \begin{align*} a_1^2 + a_2^2 + a_3^2 &= 4 \\ b_1^2 + b_2^2 + b_3^2 &= 4 \\ a_1 b_1 + a_2 b_2 + a_3 b_3 &= 0 \\ (A \times B) &= \begin{bmatrix} i & j & k \\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \\ \end{bmatrix} \\ (A \times B) &= ( a_2 b_3 - b_2 a_3, b_1 a_3 - a_1 b_3, a_1b_2 - b1_a2 ) \\ (A \times B) \times C &= \begin{vmatrix} i & j & k \\ a_2 b_3 - b_2 a_3 & b_1 a_3 - a_1 b_3 & a_1b_2 - b1_a2 \\ c_1 & c_2 & c_3 \\ \end{vmatrix} \\ (A \times B) \cdot C &= C \cdot (A \times B) = 8 \\ \begin{vmatrix} c_1 & c_2 & c_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \end{vmatrix} &= 8 \end{align*} Given the hint, I believe I am taking the wrong approach but I do not know the right approach.

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To expand a bit on J.G.’s answer, the cross product of two vectors is orthogonal to both of them, and a vanishing cross product means that one vector is a scalar multiple of the other. Putting these two facts together, $(A\times B)\times C=0$ implies that $C$ is orthogonal to both $A$ and $B$, i.e., $A\cdot C=0$. The rest of the information that you’re given in the problem is irrelevant, although it does tell us that all three vectors are nonzero.

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Since $C\parallel A\times B$, $A\cdot C=0$.

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You can see that $A \cdot B = 0 $, meaning that $A$ Is perpendicular to $B$, and from $(A\times B) \times C = 0$ It follows that $C$ is a linear combination of $A$ and $B$. Then you can use other 3 equations to find out $A\cdot C $.

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    $\begingroup$ Did you mean $C$ is a linear combination of $A$ and $\color{red}B$? $\endgroup$ – J. W. Tanner Dec 25 '19 at 13:31
  • $\begingroup$ Yeah, my bed, I’ll fix it, thanks $\endgroup$ – Uroš Kosmač Dec 25 '19 at 13:33
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    $\begingroup$ @UrošKosmač I do not understand why this is true: It follows that C is a linear combination of A and B. $\endgroup$ – Bob Dec 25 '19 at 13:35
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    $\begingroup$ Damn, I’m sorry that would be true if $(A\times B)\times C$ wasn’t 0. $C = \alpha A\times B$, becous of that equation, from this it diretcly follows that $C$ is perpendicular to $A$ and $B$ $\endgroup$ – Uroš Kosmač Dec 25 '19 at 14:12
  • $\begingroup$ Rather than making this correction in a comment, edit your answer to make it correct. $\endgroup$ – amd Dec 25 '19 at 20:36

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