# A question dealing with three vectors

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.)

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.

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.
Since $$C\parallel A\times B$$, $$A\cdot C=0$$.
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 $$A\cdot C$$.
• Did you mean $C$ is a linear combination of $A$ and $\color{red}B$? – J. W. Tanner Dec 25 '19 at 13:31
• 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 $B$ – Uroš Kosmač Dec 25 '19 at 14:12