Question

For what vectors \(\mathbf{n}\) is \((\operatorname{curl} \mathbf{F}) \cdot \mathbf{n}=0\) when \(\mathbf{F}=\langle y,-2 z,-x\rangle ?\)

Short answer

Answer: \(\mathbf{n} = \langle a, 0, c \rangle\), where \(a\) and \(c\) are any real numbers.

Step-by-step solution

Compute the curl of \(\mathbf{F}\)

To compute the curl of \(\mathbf{F}\), we will use the following formula: $$ \operatorname{curl} \mathbf{F} = (\frac{\partial(-2z)}{\partial y} - \frac{\partial(-x)}{\partial z}, \frac{\partial(y)}{\partial x}-\frac{\partial(-z)}{\partial x}, \frac{\partial(-x)}{\partial x} - \frac{\partial(y)}{\partial x}) $$ Now, find the partial derivatives and plug them into the formula for curl. $$ \operatorname{curl} \mathbf{F} = (\frac{\partial(-2z)}{\partial y} - \frac{\partial(-x)}{\partial z}, \frac{\partial(y)}{\partial x}-\frac{\partial(-z)}{\partial x}, \frac{\partial(-x)}{\partial x} - \frac{\partial(y)}{\partial x}) = (0,2,0) $$ Thus, the curl of \(\mathbf{F}\) is \((0,2,0)\).

Set the dot product of curl(\(\mathbf{F}\)) and \(\mathbf{n}\) to zero

We want to find the vectors \(\mathbf{n}\) such that the dot product of the curl of \(\mathbf{F}\) and \(\mathbf{n}\) is zero: $$ (\operatorname{curl} \mathbf{F}) \cdot \mathbf{n} = 0 $$ We found that the curl of \(\mathbf{F}\) is \((0,2,0)\). Let vector \(\mathbf{n} = \langle a,b,c \rangle\). Now, substitute these values into the dot product expression: $$ (0,2,0) \cdot \langle a,b,c \rangle = 0 $$ This simplifies to: $$ 2b = 0 $$

Find the vectors \(\mathbf{n}\)

From the previous step, we have the equation \(2b = 0\). This implies that \(b = 0\). The dot product will be zero for any vectors \(\mathbf{n} = \langle a,0,c \rangle\) where \(a\) and \(c\) can be any real numbers. So, for any vectors \(\mathbf{n} = \langle a,0,c \rangle\), where \(a\) and \(c\) are real numbers, the dot product of \((\operatorname{curl} \mathbf{F})\) and \(\mathbf{n}\) is zero.