Cross Product - Calculus 3

1. Given that $\vec{u} = \langle 3, -1, 4 \rangle$ and $\vec{v} = \langle -2, 1, 1 \rangle$, if $\vec{w}$ is a vector such that $\vec{w} \times \vec{u} = \vec{v}$, which of the following statements is necessarily true?
   Answer: No such vector $\vec{w}$ exists because $\vec{v} \cdot \vec{u} \neq 0$. For the equation $\vec{w} \times \vec{u} = \vec{v}$ to have a solution, it is necessary that $\vec{v} \cdot \vec{u} = 0$ (since the cross product is always perpendicular to both original vectors). Let's check: $\vec{v} \cdot \vec{u} = \langle -2, 1, 1 \rangle \cdot \langle 3, -1, 4 \rangle = (-2)(3) + (1)(-1) + (1)(4) = -6 - 1 + 4 = -3 \neq 0$. Since $\vec{v}$ is not perpendicular to $\vec{u}$, no such vector $\vec{w}$ can exist. Choice A is wrong because no solution exists. Choice B is wrong for the same reason. Choice D gives an incorrect reason - the relative magnitudes of $\vec{v}$ and $\vec{u}$ don't determine existence of solutions to cross product equations.
2. If $\vec{u} \times \vec{v} = \langle 4, -3, 2 \rangle$ and $||\vec{u}|| = 3$, $||\vec{v}|| = 2$, what is the acute angle $\theta$ between vectors $\vec{u}$ and $\vec{v}$?
   Answer: $\theta = \arcsin\left(\frac{\sqrt{29}}{6}\right)$ We know that $||\vec{u} \times \vec{v}|| = ||\vec{u}|| \cdot ||\vec{v}|| \sin \theta$. First, find $||\vec{u} \times \vec{v}|| = \sqrt{4^2 + (-3)^2 + 2^2} = \sqrt{16 + 9 + 4} = \sqrt{29}$. Using the cross product magnitude formula: $\sqrt{29} = 3 \cdot 2 \cdot \sin \theta = 6 \sin \theta$. Therefore, $\sin \theta = \frac{\sqrt{29}}{6}$, and $\theta = \arcsin\left(\frac{\sqrt{29}}{6}\right)$. Choice B uses arccosine instead of arcsine, which would be incorrect since we found the sine of the angle. Choice C uses $\sqrt{7}$ instead of $\sqrt{29}$, which results from incorrectly calculating the magnitude of the cross product. Choice D combines both errors, using both arccosine and the wrong magnitude.

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