\boxed{\left\langle -\dfrac{45}{14}, -\dfrac{3}{7}, \dfrac{19}{14} \right\rangle}
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- Then $v_2 = 6 + 2(-\frac{45}{14}) = \frac{84 - 90}{14} = -\frac{6}{14} = -\frac{3}{7}$, and $v_3 = 11 + 3(-\frac{45}{14}) = \frac{154 - 135}{14} = \frac{19}{14}$.
- Thus, $\mathbf{v} = \left\langle -\frac{45}{14}, -\frac{3}{7}, \frac{19}{14} \right\rangle$.
- Verification: Cross product and dot product conditions are satisfied.
- Question: In spherical coordinates, describe the surface defined by $\rho = 2\sin\phi \cos\theta$.
- Solution: Convert to Cartesian coordinates using $\rho = \sqrt{x^2 + y^2 + z^2}$, $x = \rho \sin\phi \cos\theta$, $y = \rho \sin\phi \sin\theta$, $z = \rho \cos\phi$. Multiply both sides by $\rho$:
- \rho^2 = 2\rho \sin\phi \cos\theta \implies x^2 + y^2 + z^2 = 2x.
