Help with multivariable calculus and vectors.

Question

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.

Answer 1

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 $\beta$.

Answer 2

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 $u\in\text{span}\{v,w\}$.