Key Concepts and Formulas

• Perpendicular Vectors and Dot Product: Two non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular if and only if their dot product is zero, i.e., $\vec{a} \cdot \vec{b} = 0$.
• Vector Perpendicular to Two Vectors: A vector that is perpendicular to two non-parallel vectors $\vec{u}$ and $\vec{v}$ is parallel to their cross product, $\vec{u} \times \vec{v}$.
• Unit Vector: A unit vector $\widehat{n}$ in the direction of a non-zero vector $\vec{a}$ is given by $\widehat{n} = \frac{\vec{a}}{|\vec{a}|}$.
• Scalar Triple Product: The scalar triple product of three vectors $\vec{w}$, $\vec{u}$, and $\vec{v}$ is given by $\vec{w} \cdot (\vec{u} \times \vec{v})$, which can be computed as the determinant of the matrix formed by their components.

Step-by-Step Solution

Step 1: Understand the Perpendicularity Conditions We are given that $\overrightarrow{u} \cdot \widehat{n} = 0$ and $\overrightarrow{v} \cdot \widehat{n} = 0$. This directly implies that the unit vector $\widehat{n}$ is perpendicular to both $\overrightarrow{u}$ and $\overrightarrow{v}$.

Step 2: Determine the Direction of $\widehat{n}$ using Cross Product A vector that is perpendicular to two non-parallel vectors must be parallel to their cross product. Therefore, $\widehat{n}$ is parallel to $\overrightarrow{u} \times \overrightarrow{v}$. Given $\overrightarrow{u} = \widehat{i} + \widehat{j}$ and $\overrightarrow{v} = \widehat{i} - \widehat{j}$. We calculate the cross product: $\overrightarrow{u} \times \overrightarrow{v} = (\widehat{i} + \widehat{j}) \times (\widehat{i} - \widehat{j})$ $\overrightarrow{u} \times \overrightarrow{v} = (\widehat{i} \times \widehat{i}) - (\widehat{i} \times \widehat{j}) + (\widehat{j} \times \widehat{i}) - (\widehat{j} \times \widehat{j})$ Using the properties $\widehat{i} \times \widehat{i} = \vec{0}$, $\widehat{j} \times \widehat{j} = \vec{0}$, $\widehat{i} \times \widehat{j} = \widehat{k}$, and $\widehat{j} \times \widehat{i} = -\widehat{k}$: $\overrightarrow{u} \times \overrightarrow{v} = \vec{0} - \widehat{k} - \widehat{k} - \vec{0}$ $\overrightarrow{u} \times \overrightarrow{v} = -2\widehat{k}$ This vector, $-2\widehat{k}$, points in the direction of $\widehat{n}$.

Step 3: Calculate the Magnitude of the Cross Product To find the unit vector $\widehat{n}$, we need to normalize the vector $\overrightarrow{u} \times \overrightarrow{v}$ by dividing it by its magnitude. $|\overrightarrow{u} \times \overrightarrow{v}| = |-2\widehat{k}| = \sqrt{0^2 + 0^2 + (-2)^2} = \sqrt{4} = 2$

Step 4: Form the Unit Vector $\widehat{n}$ Now we can find $\widehat{n}$ by dividing the cross product vector by its magnitude. There are two possible unit vectors, pointing in opposite directions. We can choose either one, as the final answer involves a magnitude. $\widehat{n} = \frac{\overrightarrow{u} \times \overrightarrow{v}}{|\overrightarrow{u} \times \overrightarrow{v}|} = \frac{-2\widehat{k}}{2} = -\widehat{k}$ Alternatively, we could have chosen $\widehat{n} = \widehat{k}$.

Step 5: Calculate the Dot Product $\overrightarrow{w} \cdot \widehat{n}$ We are given $\overrightarrow{w} = \widehat{i} + 2\widehat{j} + 3\widehat{k}$. We need to compute $\overrightarrow{w} \cdot \widehat{n}$. Using $\widehat{n} = -\widehat{k}$: $\overrightarrow{w} \cdot \widehat{n} = (\widehat{i} + 2\widehat{j} + 3\widehat{k}) \cdot (-\widehat{k})$ Using the properties of dot products of unit vectors ($\widehat{i} \cdot \widehat{k} = 0$, $\widehat{j} \cdot \widehat{k} = 0$, $\widehat{k} \cdot \widehat{k} = 1$): $\overrightarrow{w} \cdot \widehat{n} = (1)(0) + (2)(0) + (3)(-1)$ $\overrightarrow{w} \cdot \widehat{n} = 0 + 0 - 3 = -3$

Step 6: Calculate the Magnitude $|\overrightarrow{w} \cdot \widehat{n}|$. The problem asks for the absolute value of the dot product: $|\overrightarrow{w} \cdot \widehat{n}| = |-3| = 3$

Alternative Method using Scalar Triple Product: The expression $\overrightarrow{w} \cdot \widehat{n}$ can be written as $\overrightarrow{w} \cdot \left( \frac{\overrightarrow{u} \times \overrightarrow{v}}{|\overrightarrow{u} \times \overrightarrow{v}|} \right) = \frac{\overrightarrow{w} \cdot (\overrightarrow{u} \times \overrightarrow{v})}{|\overrightarrow{u} \times \overrightarrow{v}|}$. The numerator is the scalar triple product, which can be calculated as a determinant: $\overrightarrow{w} \cdot (\overrightarrow{u} \times \overrightarrow{v}) = \begin{vmatrix} 1 & 2 & 3 \\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{vmatrix}$ Expanding along the third column: $= 3 \left| \begin{matrix} 1 & 1 \\ 1 & -1 \end{matrix} \right| - 0 + 0 = 3((1)(-1) - (1)(1)) = 3(-1 - 1) = 3(-2) = -6$ We already found $|\overrightarrow{u} \times \overrightarrow{v}| = 2$. Therefore, $\overrightarrow{w} \cdot \widehat{n} = \frac{-6}{2} = -3$ And $|\overrightarrow{w} \cdot \widehat{n}| = |-3| = 3$.

Common Mistakes & Tips

• Forgetting to Normalize: When forming a unit vector $\widehat{n}$, ensure you divide the direction vector by its magnitude.
• Sign Errors in Cross Product: Be careful with the order of vectors in the cross product and the signs of the unit vectors ($\widehat{i} \times \widehat{j} = \widehat{k}$, but $\widehat{j} \times \widehat{i} = -\widehat{k}$).
• Magnitude vs. Value: The question asks for the magnitude $|\overrightarrow{w} \cdot \widehat{n}|$, so the sign of the dot product is irrelevant to the final answer. If $\widehat{n}$ were chosen as $\widehat{k}$, $\overrightarrow{w} \cdot \widehat{n}$ would be $3$, and its magnitude would still be $3$.

Summary

The problem requires finding a unit vector $\widehat{n}$ that is perpendicular to two given vectors $\overrightarrow{u}$ and $\overrightarrow{v}$. This direction is found using the cross product $\overrightarrow{u} \times \overrightarrow{v}$. After normalizing this cross product to obtain $\widehat{n}$, we compute its dot product with a third vector $\overrightarrow{w}$. Finally, we take the absolute value of this dot product. The scalar triple product provides an alternative and often more efficient way to calculate the required value.

The final answer is $\boxed{3}$.