Vector Cross Product Equation

If we take $V = (v_1, v_2, v_3)$ , then $A \times V = B$ yields $(2v_3 + 5v_2, -5v_1 - v_3, -2v_1 + v_2) = (90, 30, 30)$ . This result leads to the 3x3 linear system:

$\begin{bmatrix}{0} && {5} && {2} \\ {-5} && {0} && {-1} \\ {-2} && {1} && {0}\end{bmatrix} \cdot \begin{bmatrix}{v_1} \\ {v_2} \\ {v_3} \end{bmatrix} = \begin{bmatrix}{90} \\ {30} \\ {30} \end{bmatrix}$

and results in:

$\begin{bmatrix}{v_1} \\ {v_2} \\ {v_3} \end{bmatrix} = \begin{bmatrix}{-6 - \frac{k}{5}} \\ {18 - \frac{2k}{5}} \\ {k} \end{bmatrix}$ ; $k \in \mathbb{R}$

We wish to find the one vector $V \in \mathbb{R^{3}}$ that has the least magnitude, or:

$|V| = \sqrt{v_1^{2} + v_2^{2} + v_3^{2}} = \sqrt{(-6 - \frac{k}{5})^2 + (18 - \frac{2k}{5})^2 + k^2} = \sqrt{360 - 12k + \frac{6}{5}k^2}.$

Upon observation, the radicand is a concave-up parabola equal to $\frac{6}{5}(k - 5)^2 + 330$ which attains its minimum value at $k = 5$ . Hence, the minimum vector is $\boxed{V = (-7, 16, 5)}$ with a magnitude $|V| = \sqrt{330}.$

We want to find the solution(s) of

$A \times V = B$

First note that $B$ has to be orthogonal to $A$ , otherwise there is no solution. Second, we note that the solution vector $V$ is also orthogonal to $B$ . Hence, $V$ can be written in terms of two linearly independent vectors that are orthogonal to $B$ . Two such vectors are $A$ and $A \times B$ . Thus,

$V = c_1 A + c_2 A \times B$

Now we have to determine $c_1$ and $c_2$ by plugging this into the original equation.

$A \times ( c_1 A + c_2 A \times B ) = c_2 (A \times A \times B ) = B$

For the last equation, we use the following vector product identity,

$A \times B \times C = (A \cdot C) B - (A \cdot B) C$

Applying this, we get,

$A \times A \times B = (A \cdot B) A - (A \cdot A) B = -(A \cdot A) B$

because A is orthogonal to B. Hence,

$A \times V = - c_2 (A \cdot A) B = B$

which implies that

$c_2 = - \dfrac{1}{(A\cdot A)}$

Finally we have the general solution of the given vector cross product equation,

$V = c_1 A - \dfrac{1}{(A\cdot A)} (A \times B)$

where $c_1 \in \mathbb{R}$ is an arbitrary constant. Finally, we note that since $A$ and $A \times B$ are orthogonal, the minimum magnitude solution $V$ is given by,

$V^{*} = - \dfrac{1}{(A\cdot A)} (A \times B)$

with the given vectors $A$ and $B$ , we have

$A \cdot A = 1^2 + 2^2 + (-5)^2 = 30$

and

$A \times B = (1, 2, -5) \times (90, 30, 30) = 30 ( (1, 2, -5) \times (3, 1, 1) ) = 30 (7, -16, -5)$

so that,

$V^{*} = - \dfrac{ 30(7, -16, -5) }{30} = (-7, 16, 5)$

Therefore, the answer is $(-7) + 16 + 5 = 14$ .

Let V=(a,b,c). A x V = i(2c+5b) –j(c+5a) +k(b-2a), satisfy following equations, 2c+5b=90, c+5a= - 30, b-2a = 30. we can express b, and c in term of a, c= -5a – 30, b = 30 + 2a.And the magnitude of V =(a^2+ b^2+ c^2)^(1/2)= (30a^2+420^a+1800)^ (1/2). Thus we must minimize (30a^2+420a+1800). So 60a + 420 = 0, or a=-7, thus V=(-7,16, 5). The sum = 14