Key Concepts and Formulas

• Scalar Projection: The scalar projection of vector $\overrightarrow u$ onto vector $\overrightarrow v$ is $\text{Proj}_{\overrightarrow v} \overrightarrow u = \frac{\overrightarrow u \cdot \overrightarrow v}{|\overrightarrow v|}$.
• Dot Product and Perpendicularity: Two non-zero vectors $\overrightarrow u$ and $\overrightarrow v$ are perpendicular if and only if $\overrightarrow u \cdot \overrightarrow v = 0$.
• Magnitude Squared using Dot Product: For any vector $\overrightarrow x$, $|\overrightarrow x|^2 = \overrightarrow x \cdot \overrightarrow x$. This allows us to expand $|\overrightarrow u + \overrightarrow v - \overrightarrow w|^2$ using the distributive property of the dot product.

Step-by-Step Solution

Step 1: Interpret the Given Information We are given the magnitudes of three vectors: $|\overrightarrow a| = 2$, $|\overrightarrow b| = 4$, and $|\overrightarrow c| = 4$. We are also given two conditions:

1. The projection of $\overrightarrow b$ on $\overrightarrow a$ is equal to the projection of $\overrightarrow c$ on $\overrightarrow a$.
2. $\overrightarrow b$ is perpendicular to $\overrightarrow c$. Our goal is to find $|\overrightarrow a + \vec b - \overrightarrow c|$.

Step 2: Translate the Projection Condition into a Dot Product Equation The scalar projection of $\overrightarrow b$ on $\overrightarrow a$ is $\frac{\overrightarrow b \cdot \overrightarrow a}{|\overrightarrow a|}$. The scalar projection of $\overrightarrow c$ on $\overrightarrow a$ is $\frac{\overrightarrow c \cdot \overrightarrow a}{|\overrightarrow a|}$. The problem states these are equal: $\frac{\overrightarrow b \cdot \overrightarrow a}{|\overrightarrow a|} = \frac{\overrightarrow c \cdot \overrightarrow a}{|\overrightarrow a|}$ Since $|\overrightarrow a| = 2 \neq 0$, we can multiply both sides by $|\overrightarrow a|$: $\overrightarrow b \cdot \overrightarrow a = \overrightarrow c \cdot \overrightarrow a$ By the commutative property of the dot product ($\overrightarrow u \cdot \overrightarrow v = \overrightarrow v \cdot \overrightarrow u$), we can rewrite this as: $\overrightarrow a \cdot \overrightarrow b = \overrightarrow a \cdot \overrightarrow c$ This is our first key relationship.

Step 3: Translate the Perpendicularity Condition into a Dot Product Equation The condition that $\overrightarrow b$ is perpendicular to $\overrightarrow c$ means their dot product is zero: $\overrightarrow b \cdot \overrightarrow c = 0$ This is our second key relationship.

Step 4: Calculate the Square of the Desired Magnitude To find $|\overrightarrow a + \vec b - \overrightarrow c|$, we will first compute its square, $|\overrightarrow a + \vec b - \overrightarrow c|^2$. Using the property $|\overrightarrow x|^2 = \overrightarrow x \cdot \overrightarrow x$: $|\overrightarrow a + \vec b - \overrightarrow c|^2 = (\overrightarrow a + \vec b - \overrightarrow c) \cdot (\overrightarrow a + \vec b - \overrightarrow c)$ Now, we expand this dot product using the distributive property: $|\overrightarrow a + \vec b - \overrightarrow c|^2 = \overrightarrow a \cdot \overrightarrow a + \overrightarrow a \cdot \overrightarrow b - \overrightarrow a \cdot \overrightarrow c + \overrightarrow b \cdot \overrightarrow a + \overrightarrow b \cdot \overrightarrow b - \overrightarrow b \cdot \overrightarrow c - \overrightarrow c \cdot \overrightarrow a - \overrightarrow c \cdot \overrightarrow b + \overrightarrow c \cdot \overrightarrow c$ Group terms and use the commutative property ($\overrightarrow u \cdot \overrightarrow v = \overrightarrow v \cdot \overrightarrow u$): $|\overrightarrow a + \vec b - \overrightarrow c|^2 = (\overrightarrow a \cdot \overrightarrow a) + (\overrightarrow b \cdot \overrightarrow b) + (\overrightarrow c \cdot \overrightarrow c) + 2(\overrightarrow a \cdot \overrightarrow b) - 2(\overrightarrow a \cdot \overrightarrow c) - 2(\overrightarrow b \cdot \overrightarrow c)$

Step 5: Substitute Known Values and Relationships We know the magnitudes and the dot product relationships derived earlier. Let's substitute them into the expanded expression:

• $|\overrightarrow a|^2 = \overrightarrow a \cdot \overrightarrow a = 2^2 = 4$
• $|\overrightarrow b|^2 = \overrightarrow b \cdot \overrightarrow b = 4^2 = 16$
• $|\overrightarrow c|^2 = \overrightarrow c \cdot \overrightarrow c = 4^2 = 16$
• From Step 2: $\overrightarrow a \cdot \overrightarrow b = \overrightarrow a \cdot \overrightarrow c$
• From Step 3: $\overrightarrow b \cdot \overrightarrow c = 0$

Substitute these into the equation for the squared magnitude: $|\overrightarrow a + \vec b - \overrightarrow c|^2 = 4 + 16 + 16 + 2(\overrightarrow a \cdot \overrightarrow b) - 2(\overrightarrow a \cdot \overrightarrow b) - 2(0)$ Observe that the terms $2(\overrightarrow a \cdot \overrightarrow b)$ and $-2(\overrightarrow a \cdot \overrightarrow c)$ cancel out because $\overrightarrow a \cdot \overrightarrow b = \overrightarrow a \cdot \overrightarrow c$. The term $-2(\overrightarrow b \cdot \overrightarrow c)$ becomes zero because $\overrightarrow b \cdot \overrightarrow c = 0$. $|\overrightarrow a + \vec b - \overrightarrow c|^2 = 4 + 16 + 16 + 0 - 0$ $|\overrightarrow a + \vec b - \overrightarrow c|^2 = 36$

Step 6: Find the Magnitude To find the magnitude, we take the positive square root of the squared magnitude: $|\overrightarrow a + \vec b - \overrightarrow c| = \sqrt{36}$ $|\overrightarrow a + \vec b - \overrightarrow c| = 6$

Common Mistakes & Tips

• Expansion Errors: Be very careful when expanding the squared magnitude of a sum/difference of vectors. The signs of the cross terms are crucial.
• Scalar vs. Vector Projection: Ensure you understand the difference. The problem statement implies scalar projection by stating "projection of $\overrightarrow b$ on $\overrightarrow a$ is equal to...".
• Dot Product Properties: Remember that $\overrightarrow x \cdot \overrightarrow x = |\overrightarrow x|^2$ and $\overrightarrow x \cdot \overrightarrow y = \overrightarrow y \cdot \overrightarrow x$.

Summary We used the given conditions regarding vector projections and perpendicularity to establish relationships between the dot products of the vectors. Specifically, we found that $\overrightarrow a \cdot \overrightarrow b = \overrightarrow a \cdot \overrightarrow c$ and $\overrightarrow b \cdot \overrightarrow c = 0$. We then calculated the square of the magnitude of the target vector, $|\overrightarrow a + \vec b - \overrightarrow c|^2$, by expanding $(\overrightarrow a + \vec b - \overrightarrow c) \cdot (\overrightarrow a + \vec b - \overrightarrow c)$. Substituting the given magnitudes and the derived dot product relationships allowed us to simplify the expression to 36. Taking the square root yielded the final magnitude.

The final answer is $\boxed{6}$.