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I need help proving the following:

If $u \cdot(v \times w)=0$ but $v \times w \not= 0 $, show that there are constants $\lambda$ and $\mu$ such that: $u=\lambda v + \mu w$

This is for a college multivariable calculus class. I have tried expanding the expressions but it's no use. Thank you.

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    Hint: $v\times w$ is perpendicular to both $v$ and $w$, while $u\cdot(v\times w)=0$ means that $u$ is perpendicular to $v\times w$. Think about what this says about the relationship of $u$ to the plane defined by $v$ and $w$. – 2017-02-06

2 Answers 2

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Because there is a cross-product, $``\times"$, you must be using vectors in $\mathbb R^3$. Let $\mathbf u = (u_1, u_2, u_3)$, $\mathbf v = (v_1, v_2, v_3)$ and $\mathbf w = (w_1, w_2, w_3)$. Then

$$ \left| \begin{array}{ccc} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \\ \end{array} \right| = \mathbf u \cdot (\mathbf v \times \mathbf w) = 0 $$

This implies that $\mathbf u, \mathbf v$ and $\mathbf w$ are linearly dependent. Since $\mathbf v \times \mathbf w \ne \mathbf 0$, then $\mathbf v$ and $\mathbf w$ are linearly independent. Hence we must have $\mathbf u = \alpha \mathbf v + \beta \mathbf w$ for some scalars $\alpha$ and $\beta$.

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Hint:

  • $v\times w\neq 0$ means $v$ and $w$ are not scalar multiples of each other.
  • Show that $v\cdot(v\times w)=0$ and similarly $w\cdot(v\times w)=0$, this can be verified by direct computation. Thus we see that $v\times w$ is perpendicular to $\text{span}\{v,w\}$.
  • From above two steps, conclude that $\{v,w,v\times w\}$ form a set of linearly independent vectors, therefore is a basis of $\Bbb{R}^3$.
  • Write $u$ as a linear combination of $v,w,v\times v$, conclude that the coefficient of $v\times w$ must be $0$, and this implies $u\in\text{span}\{v,w\}$.