Key Concepts and Formulas

• Scalar Projection: The scalar projection of vector $\overrightarrow a$ along vector $\overrightarrow b$ is given by $\text{Proj}_{\overrightarrow b} \overrightarrow a = \frac{\overrightarrow a \cdot \overrightarrow b}{|\overrightarrow b|}$.
• Perpendicular Vectors: Two non-zero vectors $\overrightarrow a$ and $\overrightarrow b$ are perpendicular if and only if their dot product is zero, i.e., $\overrightarrow a \cdot \overrightarrow b = 0$.
• Magnitude of a Vector Sum/Difference: For any vectors $\overrightarrow A, \overrightarrow B, \overrightarrow C$, the square of the magnitude of their sum/difference is given by $|\overrightarrow A \pm \overrightarrow B \pm \overrightarrow C|^2 = (\overrightarrow A \pm \overrightarrow B \pm \overrightarrow C) \cdot (\overrightarrow A \pm \overrightarrow B \pm \overrightarrow C)$. This expands to $|\overrightarrow A|^2 + |\overrightarrow B|^2 + |\overrightarrow C|^2 \pm 2\overrightarrow A \cdot \overrightarrow B \pm 2\overrightarrow B \cdot \overrightarrow C \pm 2\overrightarrow A \cdot \overrightarrow C$, where the signs of the dot product terms depend on the signs in the original expression.

Step-by-Step Solution

Step 1: Understand and Translate Given Information

We are given the magnitudes of three vectors: $|\overrightarrow u| = 1$ $|\overrightarrow v| = 2$ $|\overrightarrow w| = 3$

We are also given two conditions:

1. The projection of $\overrightarrow v$ along $\overrightarrow u$ is equal to the projection of $\overrightarrow w$ along $\overrightarrow u$. Mathematically, this means: $\frac{\overrightarrow v \cdot \overrightarrow u}{|\overrightarrow u|} = \frac{\overrightarrow w \cdot \overrightarrow u}{|\overrightarrow u|}$
2. Vectors $\overrightarrow v$ and $\overrightarrow w$ are perpendicular to each other. Mathematically, this means: $\overrightarrow v \cdot \overrightarrow w = 0$

Step 2: Simplify the Projection Condition

From the first condition, since $|\overrightarrow u| = 1 \neq 0$, we can multiply both sides by $|\overrightarrow u|$: $\overrightarrow v \cdot \overrightarrow u = \overrightarrow w \cdot \overrightarrow u$ This implies that $\overrightarrow u \cdot \overrightarrow v = \overrightarrow u \cdot \overrightarrow w$.

Step 3: Utilize the Perpendicularity Condition

The second condition directly gives us: $\overrightarrow v \cdot \overrightarrow w = 0$

Step 4: Calculate the Square of the Magnitude of the Target Expression

We need to find $|\overrightarrow u - \overrightarrow v + \overrightarrow w|$. It is easier to first compute the square of this magnitude: $|\overrightarrow u - \overrightarrow v + \overrightarrow w|^2 = (\overrightarrow u - \overrightarrow v + \overrightarrow w) \cdot (\overrightarrow u - \overrightarrow v + \overrightarrow w)$ Expanding this dot product: $|\overrightarrow u - \overrightarrow v + \overrightarrow w|^2 = |\overrightarrow u|^2 + (-\overrightarrow v) \cdot (-\overrightarrow v) + \overrightarrow w \cdot \overrightarrow w + 2(\overrightarrow u \cdot (-\overrightarrow v)) + 2((-\overrightarrow v) \cdot \overrightarrow w) + 2(\overrightarrow w \cdot \overrightarrow u)$ $|\overrightarrow u - \overrightarrow v + \overrightarrow w|^2 = |\overrightarrow u|^2 + |\overrightarrow v|^2 + |\overrightarrow w|^2 - 2\overrightarrow u \cdot \overrightarrow v - 2\overrightarrow v \cdot \overrightarrow w + 2\overrightarrow u \cdot \overrightarrow w$

Step 5: Substitute Known Values and Derived Relationships

Now, substitute the given magnitudes and the results from Steps 2 and 3 into the expanded expression:

• $|\overrightarrow u| = 1 \implies |\overrightarrow u|^2 = 1$
• $|\overrightarrow v| = 2 \implies |\overrightarrow v|^2 = 4$
• $|\overrightarrow w| = 3 \implies |\overrightarrow w|^2 = 9$
• From Step 2: $\overrightarrow u \cdot \overrightarrow v = \overrightarrow u \cdot \overrightarrow w$
• From Step 3: $\overrightarrow v \cdot \overrightarrow w = 0$

Substituting these into the equation from Step 4: $|\overrightarrow u - \overrightarrow v + \overrightarrow w|^2 = 1 + 4 + 9 - 2\overrightarrow u \cdot \overrightarrow v - 2(0) + 2\overrightarrow u \cdot \overrightarrow w$ $|\overrightarrow u - \overrightarrow v + \overrightarrow w|^2 = 14 - 2\overrightarrow u \cdot \overrightarrow v + 2\overrightarrow u \cdot \overrightarrow w$ Using the relationship $\overrightarrow u \cdot \overrightarrow v = \overrightarrow u \cdot \overrightarrow w$ from Step 2, the terms $-2\overrightarrow u \cdot \overrightarrow v$ and $+2\overrightarrow u \cdot \overrightarrow w$ cancel each other out: $-2\overrightarrow u \cdot \overrightarrow v + 2\overrightarrow u \cdot \overrightarrow w = -2(\overrightarrow u \cdot \overrightarrow w) + 2(\overrightarrow u \cdot \overrightarrow w) = 0$ So, the expression simplifies to: $|\overrightarrow u - \overrightarrow v + \overrightarrow w|^2 = 14 + 0$ $|\overrightarrow u - \overrightarrow v + \overrightarrow w|^2 = 14$

Step 6: Find the Final Magnitude

To find the magnitude $|\overrightarrow u - \overrightarrow v + \overrightarrow w|$, we take the square root of the result from Step 5: $|\overrightarrow u - \overrightarrow v + \overrightarrow w| = \sqrt{14}$

Common Mistakes & Tips

• Projection Interpretation: Ensure you correctly distinguish between scalar and vector projections. The problem specifies "projection" which, in this context, refers to the scalar projection.
• Sign Errors: Be meticulous when expanding the square of the magnitude of a sum/difference of vectors, especially with negative signs.
• Forgetting the Square Root: The final step is to take the square root of the squared magnitude.

Summary

The problem requires applying the definitions of scalar projection and perpendicular vectors, and then utilizing the formula for the magnitude of a vector sum/difference. By translating the given conditions into dot product equations and substituting these into the expanded magnitude expression, we found that the terms involving dot products cancel out, leaving a simple numerical value for the squared magnitude. Taking the square root of this value yields the final answer.

The final answer is $\boxed{\sqrt{14}}$.